Statistics Questions

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company has 544 employees and 182 of them are college graduates Determine the proportion of all empinyees that are not college graduates 000 000
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Probability
company has 544 employees and 182 of them are college graduates Determine the proportion of all empinyees that are not college graduates 000 000
The total area under a normal distribution curve to the right of the mean is always O less than 0 5 O equal to 0 O equal to 1 Ogreater than 0 5 O equal to 0 5
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Statistics
The total area under a normal distribution curve to the right of the mean is always O less than 0 5 O equal to 0 O equal to 1 Ogreater than 0 5 O equal to 0 5
Determine the value of a such that the area between 0 and z is 0 4484 and z is positive 0a 1 68 0 0 73 0a 0 58 Oz 1 63
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Statistics
Determine the value of a such that the area between 0 and z is 0 4484 and z is positive 0a 1 68 0 0 73 0a 0 58 Oz 1 63
2 55 A random sample of 64 Male college students shows that average weight is 166 lb with a Standard deviation of 25 lb a At 5 level of significance does the sample provide enough evidence to prove that the mean weigh of all college students is more than 163 lb b Also explain the meaning of alpha and P value occurred in the content of this problem c Describe what kind of error it is for type II error in the content of this problem
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Statistics
2 55 A random sample of 64 Male college students shows that average weight is 166 lb with a Standard deviation of 25 lb a At 5 level of significance does the sample provide enough evidence to prove that the mean weigh of all college students is more than 163 lb b Also explain the meaning of alpha and P value occurred in the content of this problem c Describe what kind of error it is for type II error in the content of this problem
A researcher wanted to determine whether certain accidents were uniformly distributed over the days of the week The data show the day of the week for n 308 randomly selected accidents Is there reason to believe that the accidents occur with equal frequency with respect to the day of the week at the x 0 05 level of significance Click the icon to view the table Click here to view the Chi Square distribution table Mo P P2P7 7 H At least one proportion is different from the others OD Ho At least one proportion is different from the others 1 H P P2 P7 7 Compute the expected counts for each day of the week Day of the Week Sunday Monday Tuesday Wednesday Thursday Friday Saturday Observed Count Expected Count 45 40 31 41 42 50 59
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Statistics
A researcher wanted to determine whether certain accidents were uniformly distributed over the days of the week The data show the day of the week for n 308 randomly selected accidents Is there reason to believe that the accidents occur with equal frequency with respect to the day of the week at the x 0 05 level of significance Click the icon to view the table Click here to view the Chi Square distribution table Mo P P2P7 7 H At least one proportion is different from the others OD Ho At least one proportion is different from the others 1 H P P2 P7 7 Compute the expected counts for each day of the week Day of the Week Sunday Monday Tuesday Wednesday Thursday Friday Saturday Observed Count Expected Count 45 40 31 41 42 50 59
1 45 A random sample of 58 college students shows that average weight is 139 lbs with a Standard deviation of 20 lb a Construct a 90 percent confidence intervall for the mean weight of all college students b Does the result in part a means that 90 of students weight is in the range of your CI c If you want to be 99 confident instead and you want to control the maximal error of estimation to be 2 lbs how many more college students need to be collected
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Statistics
1 45 A random sample of 58 college students shows that average weight is 139 lbs with a Standard deviation of 20 lb a Construct a 90 percent confidence intervall for the mean weight of all college students b Does the result in part a means that 90 of students weight is in the range of your CI c If you want to be 99 confident instead and you want to control the maximal error of estimation to be 2 lbs how many more college students need to be collected
P Value 0 1874 5 2 test t 0 9 t 1 69 3
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Statistics
P Value 0 1874 5 2 test t 0 9 t 1 69 3
A company manufacturing CDs is working on a new technology A random sample of 615 Internet users were asked A you may know some CDs are being manufactured so that you can only make one copy of the CD after you purchase it Would you buy a CD with this technology or would you refuse to buy it even if it was one you would normally buy Of these users 68 responded that they would buy the CD Complete parts a and b below DI a Create a 99 confidence interval for this percentage Round to three decimal places as needed Use ascending order
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Statistics
A company manufacturing CDs is working on a new technology A random sample of 615 Internet users were asked A you may know some CDs are being manufactured so that you can only make one copy of the CD after you purchase it Would you buy a CD with this technology or would you refuse to buy it even if it was one you would normally buy Of these users 68 responded that they would buy the CD Complete parts a and b below DI a Create a 99 confidence interval for this percentage Round to three decimal places as needed Use ascending order
Use Newton s method to find all solutions of the equation correct to six decimal places 5 x 1 x 2x Ox 1 000000 x 0 887685 3 856974 x 0 799572 4 579117 x 0 000000 2 000000 x 0 345089 2 710644
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Probability
Use Newton s method to find all solutions of the equation correct to six decimal places 5 x 1 x 2x Ox 1 000000 x 0 887685 3 856974 x 0 799572 4 579117 x 0 000000 2 000000 x 0 345089 2 710644
An epidemiologist needs to estimate the proportion of residents of St Lucie county that have been infected with COVID 19 Determine the most conservative estimate of the sample size required to limit the margin of error to within 0 047 of the population proportion for a 90 confidence interval Round the solution up to the nearest whole number
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Statistics
An epidemiologist needs to estimate the proportion of residents of St Lucie county that have been infected with COVID 19 Determine the most conservative estimate of the sample size required to limit the margin of error to within 0 047 of the population proportion for a 90 confidence interval Round the solution up to the nearest whole number
Determine the critical value s for a two tailed hypothesis test for a mean with the given characteristics Round any z value solution to two decimal places Round any t value solution to four decimal places a 0 005 The sample size is 44 The population standard deviation o is known to be 8 3 Should the t or z distribution be used for the above scenario The Student s t distribution should be used The standard normal z distribution should be used The critical value s for the test are given by
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Statistics
Determine the critical value s for a two tailed hypothesis test for a mean with the given characteristics Round any z value solution to two decimal places Round any t value solution to four decimal places a 0 005 The sample size is 44 The population standard deviation o is known to be 8 3 Should the t or z distribution be used for the above scenario The Student s t distribution should be used The standard normal z distribution should be used The critical value s for the test are given by
hypothesis O 89 O 45 0 82 0 20 0 70 0 75 Which if any of the following hypotheses represent an alternative hypothesis 0 20 0 70
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Statistics
hypothesis O 89 O 45 0 82 0 20 0 70 0 75 Which if any of the following hypotheses represent an alternative hypothesis 0 20 0 70
A public health official in Okeechobee county needs to estimate the average diastolic blood pressure of all residents in Okeechobee county for a report that is being prepared for the Florida Department of Health The official randomly selected 99 Okeechobee county residents and found that the mean diastolic blood pressure of the sample was 81 millimeters of mercury mm Hg Using a 99 confidence level determine the margin of error E and a confidence interval for the mean diastolic blood pressure of all Okeechobee county residents From past research it is known that the standard deviation of the distribution of all Okeechobee county residents diastolic blood pressure is 7 mm Hg Report the confidence interval using interval notation Round solutions to two decimal places if necessary The margin of error is given by E CON A 99 confidence interval is given by
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Statistics
A public health official in Okeechobee county needs to estimate the average diastolic blood pressure of all residents in Okeechobee county for a report that is being prepared for the Florida Department of Health The official randomly selected 99 Okeechobee county residents and found that the mean diastolic blood pressure of the sample was 81 millimeters of mercury mm Hg Using a 99 confidence level determine the margin of error E and a confidence interval for the mean diastolic blood pressure of all Okeechobee county residents From past research it is known that the standard deviation of the distribution of all Okeechobee county residents diastolic blood pressure is 7 mm Hg Report the confidence interval using interval notation Round solutions to two decimal places if necessary The margin of error is given by E CON A 99 confidence interval is given by
Suppose a hypothesis test was conducted for a mean with a significance level of 2 The p value for the test was calculated to be p 0 025 Determine the correct conclusion given the result above Reject H Fail to reject Ho Reject Ho Fail to reject H Interpret the above conclusion The sample data provide sufficient evidence to reject the null hypothesis and conclude that the alternative hypothesis is likely true The sample data do not provide sufficient evidence to reject the alternative hypothessis and conclude that the alternative hypothesis is likely true The sample data provide sufficient evidence to reject the alternative hypothesis and conclude that the null hypothesis is likely true The sample data do not provide sufficient evidence to reject the null hypothesis and conclude that the null
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Statistics
Suppose a hypothesis test was conducted for a mean with a significance level of 2 The p value for the test was calculated to be p 0 025 Determine the correct conclusion given the result above Reject H Fail to reject Ho Reject Ho Fail to reject H Interpret the above conclusion The sample data provide sufficient evidence to reject the null hypothesis and conclude that the alternative hypothesis is likely true The sample data do not provide sufficient evidence to reject the alternative hypothessis and conclude that the alternative hypothesis is likely true The sample data provide sufficient evidence to reject the alternative hypothesis and conclude that the null hypothesis is likely true The sample data do not provide sufficient evidence to reject the null hypothesis and conclude that the null
A company produces ball bearings for industrial applications The bearings a manufactured to be precisely 2 75 mm in diameter Every 30 minutes a quality control officer takes a random sample of ball bearings from the assembly line and measures each to ensure that the average diameter of the sample is not significantly different than 2 75 mm Determine the null and alternative hypotheses for the above hypothesis test Ho 2 75 2 75 O Ho Ho 2 75 O Ho 2 75 H 2 75 H 2 75 H 2 75 H 2 75
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Statistics
A company produces ball bearings for industrial applications The bearings a manufactured to be precisely 2 75 mm in diameter Every 30 minutes a quality control officer takes a random sample of ball bearings from the assembly line and measures each to ensure that the average diameter of the sample is not significantly different than 2 75 mm Determine the null and alternative hypotheses for the above hypothesis test Ho 2 75 2 75 O Ho Ho 2 75 O Ho 2 75 H 2 75 H 2 75 H 2 75 H 2 75
Suppose a researcher is testing the hypothesis Ho p 0 6 versus H p 0 6 and she finds the P value to be 0 21 Explain what this means Would she reject the null hypothesis Why Choose the correct explanation below OA If the P value for a particular test statistic is 0 21 she expects results at least as extreme as the test statistic in about 21 of 100 samples if the null hypothesis is true OB If the P value for a particular test statistic is 0 21 she expects results no more extreme than the test statistic in about 21 of 100 samples if the null hypothesis is true OC If the P value for a particular test statistic is 0 21 she expects results no more extreme than the test statistic in exactly 21 of 100 samples if the null hypothesis is true OD If the P value for a particular test statistic is 0 21 she expects results at least as extreme as the test statistic in exactly 21 of 100 samples if the null hypothesis is true Choose the correct conclusion below OA Since this event is unusual she will reject the null hypothesis OB Since this event is not unusual she will reject the null hypothesis C Since this event is unusual she will not reject the null hypothesis D Since this event is not unusual she will not reject the null hypothesis
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Suppose a researcher is testing the hypothesis Ho p 0 6 versus H p 0 6 and she finds the P value to be 0 21 Explain what this means Would she reject the null hypothesis Why Choose the correct explanation below OA If the P value for a particular test statistic is 0 21 she expects results at least as extreme as the test statistic in about 21 of 100 samples if the null hypothesis is true OB If the P value for a particular test statistic is 0 21 she expects results no more extreme than the test statistic in about 21 of 100 samples if the null hypothesis is true OC If the P value for a particular test statistic is 0 21 she expects results no more extreme than the test statistic in exactly 21 of 100 samples if the null hypothesis is true OD If the P value for a particular test statistic is 0 21 she expects results at least as extreme as the test statistic in exactly 21 of 100 samples if the null hypothesis is true Choose the correct conclusion below OA Since this event is unusual she will reject the null hypothesis OB Since this event is not unusual she will reject the null hypothesis C Since this event is unusual she will not reject the null hypothesis D Since this event is not unusual she will not reject the null hypothesis
Complete parts a through c below a Determine the critical value s for a right tailed test of a population mean at the x 0 01 level of significance with 10 degrees of freedom b Determine the critical value s for a left tailed test of a population mean at the 0 01 level of significance based on a sample size of n 20 c Determine the critical value s for a two tailed test of a population mean at the x 0 05 level of significance based on a sample size of n 18 Click here to view the t Distribution Area in Right Tail a tcrit b tcrit e t Round to three decimal places as needed Round to three decimal places as needed Bound to three decimal places as needed
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Statistics
Complete parts a through c below a Determine the critical value s for a right tailed test of a population mean at the x 0 01 level of significance with 10 degrees of freedom b Determine the critical value s for a left tailed test of a population mean at the 0 01 level of significance based on a sample size of n 20 c Determine the critical value s for a two tailed test of a population mean at the x 0 05 level of significance based on a sample size of n 18 Click here to view the t Distribution Area in Right Tail a tcrit b tcrit e t Round to three decimal places as needed Round to three decimal places as needed Bound to three decimal places as needed
Explain what a P value is What is the criterion for rejecting the null hypothesis using the P value approach Explain what a P value is Choose the correct answer below OA A P value is the number of standard deviations that the observed proportion is from the proportion stated in the null hypothesis OB A P value is the value used to designate the area in either the left or right tail of the normal curve OC A P value is the probability of observing a sample statistic as extreme or more extreme than the one observed under the assumption that the statement in the null hypothesis is true What is the criterion for rejecting the null hypothesis using the P value approach Choose the correct answer below OA If P value a reject the null hypothesis OB If P value z for a left tailed test or if P value z for a right tailed test or if P value Z 2 or P value Za 2 for a two tailed test then reject the null hypothesis OS If P value a reiect the null hypothesis
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Statistics
Explain what a P value is What is the criterion for rejecting the null hypothesis using the P value approach Explain what a P value is Choose the correct answer below OA A P value is the number of standard deviations that the observed proportion is from the proportion stated in the null hypothesis OB A P value is the value used to designate the area in either the left or right tail of the normal curve OC A P value is the probability of observing a sample statistic as extreme or more extreme than the one observed under the assumption that the statement in the null hypothesis is true What is the criterion for rejecting the null hypothesis using the P value approach Choose the correct answer below OA If P value a reject the null hypothesis OB If P value z for a left tailed test or if P value z for a right tailed test or if P value Z 2 or P value Za 2 for a two tailed test then reject the null hypothesis OS If P value a reiect the null hypothesis
autosave questio Consider a train that runs from Seattle to Los Angeles The mean travel time from one stop to the next is 130 minutes with a standard deviation of 114 minutes The mean distance traveled from one stop to the next is 108 miles with a standard deviation of 100 miles The correlation between travel time and distance is 0 631 Travel Time minutes 360 300 240 180 120 60 100 Assignment Responses submit dep 300 200 Distance miles a Write the equation of the regression line for predicting travel time Let y represent travel time minutes and let x represent distance traveled in miles b Interpret the slope in this context For each mile increase in distance traveled we would expect travel time to increase on average by Interpret the intercept in this context When the distance traveled is 0 miles the travel time is expected to be on average minutes minutes c Calculate R2 of the regression line for predicting travel time from distance traveled for the train Round your answer to three decimal places R Interpret R2 in the context of the application Give your answer as a percent Round your answer to one decimal place Approximately of the variation in travel time is accounted for by the model The distance between Santa Barbara and Los Angeles is 103 miles Use the model to estimate the time in minutes it takes for the train to travel between these two cities Round your answer to the
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Statistics
autosave questio Consider a train that runs from Seattle to Los Angeles The mean travel time from one stop to the next is 130 minutes with a standard deviation of 114 minutes The mean distance traveled from one stop to the next is 108 miles with a standard deviation of 100 miles The correlation between travel time and distance is 0 631 Travel Time minutes 360 300 240 180 120 60 100 Assignment Responses submit dep 300 200 Distance miles a Write the equation of the regression line for predicting travel time Let y represent travel time minutes and let x represent distance traveled in miles b Interpret the slope in this context For each mile increase in distance traveled we would expect travel time to increase on average by Interpret the intercept in this context When the distance traveled is 0 miles the travel time is expected to be on average minutes minutes c Calculate R2 of the regression line for predicting travel time from distance traveled for the train Round your answer to three decimal places R Interpret R2 in the context of the application Give your answer as a percent Round your answer to one decimal place Approximately of the variation in travel time is accounted for by the model The distance between Santa Barbara and Los Angeles is 103 miles Use the model to estimate the time in minutes it takes for the train to travel between these two cities Round your answer to the
Let A and B be events in a sample space S and let C S AU B Suppose P A 0 2 P B 0 5 and P A n B 0 1 Find each of the following a P AUB 0 6 b P C 0 4 c P AC 0 8 d P AC n BC 0 4 e P AC UBC 0 9 f P BC n C 103
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Probability
Let A and B be events in a sample space S and let C S AU B Suppose P A 0 2 P B 0 5 and P A n B 0 1 Find each of the following a P AUB 0 6 b P C 0 4 c P AC 0 8 d P AC n BC 0 4 e P AC UBC 0 9 f P BC n C 103
An airline company recently boasted about its 72 customer satisfaction rate A corporate watchdog group believes that this percentage is greatly exaggerated and the real percentage of satisfied customers is lower than 72 The group randomly surveyed 922 recent customers of this airline and found that only 648 reported being satisfied with the airline Using a significance level of 5 test the hypothesis that the proportion of satisfied customers of this airline is less than 72 Use the critical value method State the null and alternative hypothesis for this test Ho V H 7 Determine if this test is left tailed right tailed or two tailed Oright tailed Otwo tailed Oleft tailed Should the standard normal 2 distribution or Student s t distribution be used for this test O The standard normal 2 distribution should be used O The Student s t distribution should be used Determine the critical value s for this hypothesis test Round the solution s to two decimal places If more than one critical value exists enter the solutions using a comma separated list Determine the test statistic Round the solution to two decimal places Determine the appropriate conclusion for this hypothesis test O The sample data do not provide sufficient evidence to reject the null hypothesis that the proportion of all customers of this airline that are satisfied is 0 72 and thus we conclude that the proportion of all customers of this airline that are satisfied is likely 0 72 O The sample data provide sufficient evidence to reject the alternative hypothesis that the proportion of all customers of this airline that are satisfied is less than 0 72 and thus we conclude that the proportion of all customers of this airline that are satisfied is likely 0 72 O The sample data do not provide sufficient evidence to reject the alternative hypothesis that the proportion of all customers of this airtine that are satisfied is less than 0 72 and thus we conclude that the proportion of all customers of this airline that are satisfied is likely less than 0 72 O The sample data provide sufficient evidence to reject the null hypothesis that the proportion of all customers of this airline that are satisfied is 0 72 and thus we conclude that the proportion of all customers of this airline that are satisfied is likely less than 0 72
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Statistics
An airline company recently boasted about its 72 customer satisfaction rate A corporate watchdog group believes that this percentage is greatly exaggerated and the real percentage of satisfied customers is lower than 72 The group randomly surveyed 922 recent customers of this airline and found that only 648 reported being satisfied with the airline Using a significance level of 5 test the hypothesis that the proportion of satisfied customers of this airline is less than 72 Use the critical value method State the null and alternative hypothesis for this test Ho V H 7 Determine if this test is left tailed right tailed or two tailed Oright tailed Otwo tailed Oleft tailed Should the standard normal 2 distribution or Student s t distribution be used for this test O The standard normal 2 distribution should be used O The Student s t distribution should be used Determine the critical value s for this hypothesis test Round the solution s to two decimal places If more than one critical value exists enter the solutions using a comma separated list Determine the test statistic Round the solution to two decimal places Determine the appropriate conclusion for this hypothesis test O The sample data do not provide sufficient evidence to reject the null hypothesis that the proportion of all customers of this airline that are satisfied is 0 72 and thus we conclude that the proportion of all customers of this airline that are satisfied is likely 0 72 O The sample data provide sufficient evidence to reject the alternative hypothesis that the proportion of all customers of this airline that are satisfied is less than 0 72 and thus we conclude that the proportion of all customers of this airline that are satisfied is likely 0 72 O The sample data do not provide sufficient evidence to reject the alternative hypothesis that the proportion of all customers of this airtine that are satisfied is less than 0 72 and thus we conclude that the proportion of all customers of this airline that are satisfied is likely less than 0 72 O The sample data provide sufficient evidence to reject the null hypothesis that the proportion of all customers of this airline that are satisfied is 0 72 and thus we conclude that the proportion of all customers of this airline that are satisfied is likely less than 0 72
Consider the diamond prices standardized by weight for diamonds with weights 0 99 carat and 1 carat See the table for summary statistics and then construct a 95 confidence interval for the average difference between the standardized prices of 0 99 and 1 carat diamonds You may assume the conditions for inference are met Round your answers to two decimal places Mean SD n 0 99 carat 44 51 12 31 23 1 carat 55 81 15 11 23
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Statistics
Consider the diamond prices standardized by weight for diamonds with weights 0 99 carat and 1 carat See the table for summary statistics and then construct a 95 confidence interval for the average difference between the standardized prices of 0 99 and 1 carat diamonds You may assume the conditions for inference are met Round your answers to two decimal places Mean SD n 0 99 carat 44 51 12 31 23 1 carat 55 81 15 11 23
According to a study conducted three years ago by the Organisation for Economic Cooperation and Development OECD the average length of stay in a hospital was 3 6 days in Australia A researcher in Australia suspects that the average length of stay in a hospital has actually increased in Australia since the study was released by the OECD three years ago To test her suspicion the researcher selected a random sample of 165 hospital patients in Australia and calculated the average length of hospital stay The mean of the sample was found to be 3 69 days Using a significance level of 2 test the hypothesis that the mean length of stay in a hospital in Australia is more than 3 6 days Assume that the standard deviation of the lengths of all hospital stays in Australia is known to be 0 46 days Use the p value method State the null and alternative hypothesis for this test Ho V H V Determine if this test is left tailed right tailed or two tailed Otwo tailed Oright tailed Oleft tailed Should the standard normal z distribution or Student s t distribution be used for this test O The standard normal z distribution should be used O The Student s t distribution should be used Determine the test statistic for the hypothesis test Round the solution to two decimal places Determine the p value for the hypothesis test Round the solution to four decimal places Determine the appropriate conclusion for this hypothesis test O The sample data provide sufficient evidence to reject the alternative hypothesis that the average length of stay in a hospital in Australia is greater than 3 6 days and thus we conclude that the average length of stay in a hospital in Australia is likely 3 6 days The sample data do not provide sufficient evidence to reject the alternative hypothesis that the average length of stay in a hospital in Australia is greater than 3 6 days and thus we conclude that the average length of stay in a hospital in Australia is likely greater than 3 6 days The sample data provide sufficient evidence to reject the null hypothesis that the average length of stay in a hospital in Australia is 3 6 days and thus we conclude that the average length of stay in a hospital in Australia is likely longer than 3 6 days The sample data do not provide sufficient evidence to reject the null hypothesis that the average length of stay in a hospital in Australia is 3 6 days and thus we conclude that the average length of stay in a hospital in Australia is likely 3 6 days
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Statistics
According to a study conducted three years ago by the Organisation for Economic Cooperation and Development OECD the average length of stay in a hospital was 3 6 days in Australia A researcher in Australia suspects that the average length of stay in a hospital has actually increased in Australia since the study was released by the OECD three years ago To test her suspicion the researcher selected a random sample of 165 hospital patients in Australia and calculated the average length of hospital stay The mean of the sample was found to be 3 69 days Using a significance level of 2 test the hypothesis that the mean length of stay in a hospital in Australia is more than 3 6 days Assume that the standard deviation of the lengths of all hospital stays in Australia is known to be 0 46 days Use the p value method State the null and alternative hypothesis for this test Ho V H V Determine if this test is left tailed right tailed or two tailed Otwo tailed Oright tailed Oleft tailed Should the standard normal z distribution or Student s t distribution be used for this test O The standard normal z distribution should be used O The Student s t distribution should be used Determine the test statistic for the hypothesis test Round the solution to two decimal places Determine the p value for the hypothesis test Round the solution to four decimal places Determine the appropriate conclusion for this hypothesis test O The sample data provide sufficient evidence to reject the alternative hypothesis that the average length of stay in a hospital in Australia is greater than 3 6 days and thus we conclude that the average length of stay in a hospital in Australia is likely 3 6 days The sample data do not provide sufficient evidence to reject the alternative hypothesis that the average length of stay in a hospital in Australia is greater than 3 6 days and thus we conclude that the average length of stay in a hospital in Australia is likely greater than 3 6 days The sample data provide sufficient evidence to reject the null hypothesis that the average length of stay in a hospital in Australia is 3 6 days and thus we conclude that the average length of stay in a hospital in Australia is likely longer than 3 6 days The sample data do not provide sufficient evidence to reject the null hypothesis that the average length of stay in a hospital in Australia is 3 6 days and thus we conclude that the average length of stay in a hospital in Australia is likely 3 6 days
A is the value of a statistic that estimates the value of a parameter Complete the statement below A is the value of a statistic that estimates the value of a parameter
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Statistics
A is the value of a statistic that estimates the value of a parameter Complete the statement below A is the value of a statistic that estimates the value of a parameter
A national survey of 1500 adult citizens of a nation found that 18 dreaded Valentine s Day The margin of error for the survey was 11 7 percentage points with 85 confidence Explain what this means Which statement below is the best explanation OA In 85 of samples of adult citizens of the nation the proportion that dreaded Valentine s Day is between 0 063 and 0 297 OB There is 85 confidence that 18 of the adult citizens of the nation dreaded Valentine s Day OC There is 73 3 to 96 7 confidence that 18 of the adult citizens of the nation dreaded Valentine s Day OD There is 85 confidence that the proportion of the adult citizens of the nation that dreaded Valentine s Day is between 0 063 and 0 297
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Statistics
A national survey of 1500 adult citizens of a nation found that 18 dreaded Valentine s Day The margin of error for the survey was 11 7 percentage points with 85 confidence Explain what this means Which statement below is the best explanation OA In 85 of samples of adult citizens of the nation the proportion that dreaded Valentine s Day is between 0 063 and 0 297 OB There is 85 confidence that 18 of the adult citizens of the nation dreaded Valentine s Day OC There is 73 3 to 96 7 confidence that 18 of the adult citizens of the nation dreaded Valentine s Day OD There is 85 confidence that the proportion of the adult citizens of the nation that dreaded Valentine s Day is between 0 063 and 0 297
What type of variable is required to construct a confidence interval for a population proportion Choose the correct answer below OA Qualitative with 3 or more possible outcomes OB Qualitative with 2 possible outcomes OC Quantitative OD Continuous O E Discrete OF Qualitative with any number of possible outcomes
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Statistics
What type of variable is required to construct a confidence interval for a population proportion Choose the correct answer below OA Qualitative with 3 or more possible outcomes OB Qualitative with 2 possible outcomes OC Quantitative OD Continuous O E Discrete OF Qualitative with any number of possible outcomes
Explain why the t distribution has less spread as the number of degrees of freedom increases Choose the correct answer below OA The t distribution has less spread as the degrees of freedom increase because as n increases less informati is known about o by the law of large numbers OB The t distribution has less spread as the degrees of freedom increase because for large values of n n 30 the t distribution and the normal distribution are the same OC The t distribution has less spread as the degrees of freedom increase because as n increases s becomes closer to o by the law of large numbers O D The t distribution has less spread as the degrees of freedom increase because the variability introduced into the t statistic becomes greater as n increases
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Statistics
Explain why the t distribution has less spread as the number of degrees of freedom increases Choose the correct answer below OA The t distribution has less spread as the degrees of freedom increase because as n increases less informati is known about o by the law of large numbers OB The t distribution has less spread as the degrees of freedom increase because for large values of n n 30 the t distribution and the normal distribution are the same OC The t distribution has less spread as the degrees of freedom increase because as n increases s becomes closer to o by the law of large numbers O D The t distribution has less spread as the degrees of freedom increase because the variability introduced into the t statistic becomes greater as n increases
High cholesterol is one of the major risk factors for heart disease A total cholesterol reading above 250 mg dL is considered high A physician is concerned that a patient may be at risk for high cholesterol In order to accurately measure the patient s average cholesterol level the physician instructs the patient to come into the physician s office once a week for 12 weeks for a blood lipid screening The results of the 12 measurements are listed below 25 252 254 256 246 250 243 260 258 254 241 249 243 Use the critical value method to test the hypothesis that the patient s cholesterol level is greater than 250 mg dL using a 0 025 Assume that the distribution of all cholesterol measurements from this patient is approximately normally distributed State the null and alternative hypothesis for this test Ho V H V Determine if this test is left tailed right tailed or two tailed Oright tailed Oleft tailed Otwo tailed Should the standard normal z distribution or Student s t distribution be used for this test O The Student s t distribution should be used O The standard normal z distribution should be used Determine the critical value s for this hypothesis test Round the solution s to four decimal places If more than one critical value exists enter the solutions using a comma separated list Determine the test statistic Round the solution to four decimal places Determine the appropriate conclusion for this hypothesis test The sample data provide sufficient evidence to reject the null hypothesis that the mean cholesterol level of the patient is not more than 250 mg dL and thus we conclude that it is likely the patient has sustained high cholesterol O The sample data do not provide sufficient evidence to reject the alternative hypothesis that the patient has sustained high cholesterol and thus we conclude that the patient s cholesterol level is likely above 250 mg dL The sample data do not provide sufficient evidence to reject the null hypothesis that the mean cholesterol level of the patient is not more than 250 mg dL and thus we conclude that it is likely the patient does not currently have sustained high cholesterol O The sample data provide sufficient evidence to reject the alternative hypothesis that the patient has sustained high cholesterol and thus we conclude that the patient s cholesterol level is likely at or below 250 mg dL
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Statistics
High cholesterol is one of the major risk factors for heart disease A total cholesterol reading above 250 mg dL is considered high A physician is concerned that a patient may be at risk for high cholesterol In order to accurately measure the patient s average cholesterol level the physician instructs the patient to come into the physician s office once a week for 12 weeks for a blood lipid screening The results of the 12 measurements are listed below 25 252 254 256 246 250 243 260 258 254 241 249 243 Use the critical value method to test the hypothesis that the patient s cholesterol level is greater than 250 mg dL using a 0 025 Assume that the distribution of all cholesterol measurements from this patient is approximately normally distributed State the null and alternative hypothesis for this test Ho V H V Determine if this test is left tailed right tailed or two tailed Oright tailed Oleft tailed Otwo tailed Should the standard normal z distribution or Student s t distribution be used for this test O The Student s t distribution should be used O The standard normal z distribution should be used Determine the critical value s for this hypothesis test Round the solution s to four decimal places If more than one critical value exists enter the solutions using a comma separated list Determine the test statistic Round the solution to four decimal places Determine the appropriate conclusion for this hypothesis test The sample data provide sufficient evidence to reject the null hypothesis that the mean cholesterol level of the patient is not more than 250 mg dL and thus we conclude that it is likely the patient has sustained high cholesterol O The sample data do not provide sufficient evidence to reject the alternative hypothesis that the patient has sustained high cholesterol and thus we conclude that the patient s cholesterol level is likely above 250 mg dL The sample data do not provide sufficient evidence to reject the null hypothesis that the mean cholesterol level of the patient is not more than 250 mg dL and thus we conclude that it is likely the patient does not currently have sustained high cholesterol O The sample data provide sufficient evidence to reject the alternative hypothesis that the patient has sustained high cholesterol and thus we conclude that the patient s cholesterol level is likely at or below 250 mg dL
alculate the critical z value s for each of the given hypothesis test scenarios below If multiple critical alues exist for a single scenario enter the solutions using a comma separated list Round z values to two ecimal places Find the critical z value s for a left tailed test of hypothesis for a proportion with a sample size of 148 and a significance level of 2 5 Z 1 96 x Find the critical z value s for a right tailed test of hypothesis for a mean assuming the population standard deviation is known with a sample size of 131 and let a 0 0005 z 3 29 o Find the critical z value s for a two tailed test of hypothesis for a mean assuming the population standard deviation is known with a sample size of 42 and a significance level of 5 Z 1 96 0 Find the critical z value s for a two tailed test of hypothesis for a mean assuming the population standard deviation is known with a sample size of 66 and let a 0 2 Z 1 28 0
Statistics
Statistics
alculate the critical z value s for each of the given hypothesis test scenarios below If multiple critical alues exist for a single scenario enter the solutions using a comma separated list Round z values to two ecimal places Find the critical z value s for a left tailed test of hypothesis for a proportion with a sample size of 148 and a significance level of 2 5 Z 1 96 x Find the critical z value s for a right tailed test of hypothesis for a mean assuming the population standard deviation is known with a sample size of 131 and let a 0 0005 z 3 29 o Find the critical z value s for a two tailed test of hypothesis for a mean assuming the population standard deviation is known with a sample size of 42 and a significance level of 5 Z 1 96 0 Find the critical z value s for a two tailed test of hypothesis for a mean assuming the population standard deviation is known with a sample size of 66 and let a 0 2 Z 1 28 0
23 of all American adults lived in a middle class household This year an economist collected data from a random sample of 1475 American adults in order to determine if the percent of American adults who live in a middle class household is lower than 23 State the hypotheses and explain the possible Type 1 and Type 2 errors Determine the null and alternative hypotheses Ho HA A Type I error in the context of this problem would be Select an answer A Type II error in the context of this problem would be Select an answer
Statistics
Statistics
23 of all American adults lived in a middle class household This year an economist collected data from a random sample of 1475 American adults in order to determine if the percent of American adults who live in a middle class household is lower than 23 State the hypotheses and explain the possible Type 1 and Type 2 errors Determine the null and alternative hypotheses Ho HA A Type I error in the context of this problem would be Select an answer A Type II error in the context of this problem would be Select an answer
During routine conversations the CEO of a new start up reports that 10 of adults between the ages of 21 and 39 will purchase her new product Hearing this some investors decide to conduct large scale study hoping to estimate the proportion to within 2 with 95 confidence How many randomly selected adults between the ages of 21 and 39 must they survey The number of adults that should be surveyed is Round up to the nearest whole number CCO
Statistics
Statistics
During routine conversations the CEO of a new start up reports that 10 of adults between the ages of 21 and 39 will purchase her new product Hearing this some investors decide to conduct large scale study hoping to estimate the proportion to within 2 with 95 confidence How many randomly selected adults between the ages of 21 and 39 must they survey The number of adults that should be surveyed is Round up to the nearest whole number CCO
Determine a the x test statistic b the degrees of freedom c the critical value using a 0 05 and d test the hypothesis at the 0 05 level of significance 1 Ho PA PB PC PD 4 H At least one of the proportions is different from the others a The test statistic is Type an exact answer Outcome Observed Expected A B C 89 100 96 111 100 100 C 104 100
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Statistics
Determine a the x test statistic b the degrees of freedom c the critical value using a 0 05 and d test the hypothesis at the 0 05 level of significance 1 Ho PA PB PC PD 4 H At least one of the proportions is different from the others a The test statistic is Type an exact answer Outcome Observed Expected A B C 89 100 96 111 100 100 C 104 100
A book claims that more hockey players are born in January through March than in October through December The following data show the number of players selected in a draft of new players for a hockey league according to their birth month Is there evidence to suggest that hockey players birthdates are not uniformly distributed throughout the year Use the level of significance a 0 01 Click the icon to view the table Click the icon to view the chi square table of critical values B Ho The distribution of hockey players birth months is uniformly distributed H The distribution of hockey players birth months is not uniformly distributed O C Ho The distribution of hockey players birth months is uniformly distributed H More hockey players are born in January March than October December O D Ho The distribution of hockey players birth months is not uniformly distributed H The distribution of hockey players birth months is uniformly distributed Compute the expected counts for each birth month The total number of hockey players is 174 Observed Count Expected Count 64 58 28 Birth Month January March April June July September October December 24 Round to two decimal places as needed
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Probability
A book claims that more hockey players are born in January through March than in October through December The following data show the number of players selected in a draft of new players for a hockey league according to their birth month Is there evidence to suggest that hockey players birthdates are not uniformly distributed throughout the year Use the level of significance a 0 01 Click the icon to view the table Click the icon to view the chi square table of critical values B Ho The distribution of hockey players birth months is uniformly distributed H The distribution of hockey players birth months is not uniformly distributed O C Ho The distribution of hockey players birth months is uniformly distributed H More hockey players are born in January March than October December O D Ho The distribution of hockey players birth months is not uniformly distributed H The distribution of hockey players birth months is uniformly distributed Compute the expected counts for each birth month The total number of hockey players is 174 Observed Count Expected Count 64 58 28 Birth Month January March April June July September October December 24 Round to two decimal places as needed
A manufacturer of colored candies states that 13 of the candies in a bag should be brown 14 yellow 13 red 24 blue 20 orange and 16 green A student randomly selected a bag of colored candies He counted the number of candies of each color and obtained the results shown in the table Test whether the bag of colored candies follows the distribution stated above at a 0 05 level of significance Using the level of significance x 0 05 test whether the color distribution is the same Click here to view the table Click here to view the table of critical values of the chi square distribution Determine the null and alternative hypotheses Choose the correct answer below O A Ho The distribution of colors is not the same as stated by the manufacturer H The distribution of colors is the same as stated by the manufacturer OB Ho The distribution of colors is at most as uniform as stated by the manufacturer H The distribution of colors is more uniform than stated by the manufacturer OC Ho The distribution of colors is at least as uniform as stated by the manufacturer H The distribution of colors is less uniform than stated by the manufacturer O D Ho The distribution of colors is the same as stated by the manufacturer H The distribution of colors is not the same as stated by the manufacturer
Statistics
Statistics
A manufacturer of colored candies states that 13 of the candies in a bag should be brown 14 yellow 13 red 24 blue 20 orange and 16 green A student randomly selected a bag of colored candies He counted the number of candies of each color and obtained the results shown in the table Test whether the bag of colored candies follows the distribution stated above at a 0 05 level of significance Using the level of significance x 0 05 test whether the color distribution is the same Click here to view the table Click here to view the table of critical values of the chi square distribution Determine the null and alternative hypotheses Choose the correct answer below O A Ho The distribution of colors is not the same as stated by the manufacturer H The distribution of colors is the same as stated by the manufacturer OB Ho The distribution of colors is at most as uniform as stated by the manufacturer H The distribution of colors is more uniform than stated by the manufacturer OC Ho The distribution of colors is at least as uniform as stated by the manufacturer H The distribution of colors is less uniform than stated by the manufacturer O D Ho The distribution of colors is the same as stated by the manufacturer H The distribution of colors is not the same as stated by the manufacturer
If the expected count of a category is less than 1 what can be done to the categories so that a goodness of fit test can still be performed Choose the correct answer below OA Two of the categories can be combined or the sample size can be increased OB The assumed probabilities can be adjusted so that none of the expected counts are less than 1 OC Nothing needs to be done The expected count can be approximated as 1 OD Nothing can be done The test cannot be performed
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Probability
If the expected count of a category is less than 1 what can be done to the categories so that a goodness of fit test can still be performed Choose the correct answer below OA Two of the categories can be combined or the sample size can be increased OB The assumed probabilities can be adjusted so that none of the expected counts are less than 1 OC Nothing needs to be done The expected count can be approximated as 1 OD Nothing can be done The test cannot be performed
State the requirements to perform a goodness of fit test Choose the correct answer below Select all that apply A at least 80 of expected frequencies 25 B at least 90 of expected frequencies 10 C at least 80 of expected frequencies 210 D all expected frequencies 25 E all expected frequencies 21 F at least 90 of expected frequencies 25
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Statistics
State the requirements to perform a goodness of fit test Choose the correct answer below Select all that apply A at least 80 of expected frequencies 25 B at least 90 of expected frequencies 10 C at least 80 of expected frequencies 210 D all expected frequencies 25 E all expected frequencies 21 F at least 90 of expected frequencies 25
Explain why chi square goodness of fit tests are always right tailed Choose the correct answer below O A The chi square goodness of fit tests are always right tailed because negative values of the test statistic are important OB The chi square goodness of fit tests are always right tailed by convention A left tail test will always yield the same results OC The chi square goodness of fit tests are always right tailed because the chi square distribution is skewed to right O D The chi square goodness of fit tests are always right tailed because the numerator in the test statistic is squared making every test statistic other than a perfect fit positive
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Probability
Explain why chi square goodness of fit tests are always right tailed Choose the correct answer below O A The chi square goodness of fit tests are always right tailed because negative values of the test statistic are important OB The chi square goodness of fit tests are always right tailed by convention A left tail test will always yield the same results OC The chi square goodness of fit tests are always right tailed because the chi square distribution is skewed to right O D The chi square goodness of fit tests are always right tailed because the numerator in the test statistic is squared making every test statistic other than a perfect fit positive
A traffic safety company publishes reports about motorcycle fatalities and helmet use In the first accompanying data table the distribution shows the proportion of fatalities by location of injury for motorcycle accidents The second data table shows the location of injury and fatalities for 2046 riders not wearing a helmet Complete parts a and b below Click the icon to view the tables Click the icon to view the chi square table of critical values a Does the distribution of fatal injuries for riders not wearing a helmet follow the distribution for all riders Use 0 10 level of significance What are the null and alternative hypotheses Ho The distribution of fatal injuries for riders not wearing a helmet riders H The distribution of fatal injuries for riders not wearing a helmet riders the same distribution for all other the same distribution for all other
Statistics
Statistics
A traffic safety company publishes reports about motorcycle fatalities and helmet use In the first accompanying data table the distribution shows the proportion of fatalities by location of injury for motorcycle accidents The second data table shows the location of injury and fatalities for 2046 riders not wearing a helmet Complete parts a and b below Click the icon to view the tables Click the icon to view the chi square table of critical values a Does the distribution of fatal injuries for riders not wearing a helmet follow the distribution for all riders Use 0 10 level of significance What are the null and alternative hypotheses Ho The distribution of fatal injuries for riders not wearing a helmet riders H The distribution of fatal injuries for riders not wearing a helmet riders the same distribution for all other the same distribution for all other
An environmental agency worries that many cars may be violating clean air emissions standards The agency hopes to check a sample of vehicles in order to estimate that percentage with a marg of error of 4 and 99 confidence To gauge the size of the problem the agency first picks 40 cars and finds 10 with faulty emissions systems How many should be sampled for a full investigatio The environmental agency should sample at least Round up to the nearest whole number vehicles
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Probability
An environmental agency worries that many cars may be violating clean air emissions standards The agency hopes to check a sample of vehicles in order to estimate that percentage with a marg of error of 4 and 99 confidence To gauge the size of the problem the agency first picks 40 cars and finds 10 with faulty emissions systems How many should be sampled for a full investigatio The environmental agency should sample at least Round up to the nearest whole number vehicles
Suppose there are n independent trials of an experiment with k 3 mutually exclusive outcomes where p represents the probability of observing the ith outcome What would be the formula of an expected count in this situation Choose the correct answer below GREENB OA The expected counts for each possible outcome are given by E n OB The expected counts for each possible outcome are given by E P n Oc The expected counts for each possible outcome are given by E Pi OD The expected counts for each possible outcome are given by E np
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Statistics
Suppose there are n independent trials of an experiment with k 3 mutually exclusive outcomes where p represents the probability of observing the ith outcome What would be the formula of an expected count in this situation Choose the correct answer below GREENB OA The expected counts for each possible outcome are given by E n OB The expected counts for each possible outcome are given by E P n Oc The expected counts for each possible outcome are given by E Pi OD The expected counts for each possible outcome are given by E np
Determine the expected count for each outcome The expected count for outcome 1 is Round to two decimal places as needed n 594 i P 1 2 3 0 18 0 41 0 21 C
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Probability
Determine the expected count for each outcome The expected count for outcome 1 is Round to two decimal places as needed n 594 i P 1 2 3 0 18 0 41 0 21 C
and District of Columbia are also included Percent Own Their Home 75 70 65 60 55 50 45 40 40 ramilies who own their home vs the percent of the population living in urban areas There are 52 observations each corresponding to a state in the US Puerto Rico 50 60 70 80 90 100 Percent Urban Population a Describe the relationship between the percent of families who own their home and the percent of the population living in urban areas Select all that apply O The relationship appears to be perfectly linear O The relationship appears to be negative O There is one potential outlier the area where 100 of the population is urban The relationship appears to be moderate There are no outliers The relationship appears to be somewhat linear O The relationship appears to be weak O The relationship appears to be positive
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Statistics
and District of Columbia are also included Percent Own Their Home 75 70 65 60 55 50 45 40 40 ramilies who own their home vs the percent of the population living in urban areas There are 52 observations each corresponding to a state in the US Puerto Rico 50 60 70 80 90 100 Percent Urban Population a Describe the relationship between the percent of families who own their home and the percent of the population living in urban areas Select all that apply O The relationship appears to be perfectly linear O The relationship appears to be negative O There is one potential outlier the area where 100 of the population is urban The relationship appears to be moderate There are no outliers The relationship appears to be somewhat linear O The relationship appears to be weak O The relationship appears to be positive
It s believed that as many as 23 of adults over 50 never graduated from high school We wish to see if this percentage is the same among the 25 to 30 age group What sample size would a to increase our confidence level to 95 while reducing the margin of error to only 5 n Round up to the nearest integer
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Probability
It s believed that as many as 23 of adults over 50 never graduated from high school We wish to see if this percentage is the same among the 25 to 30 age group What sample size would a to increase our confidence level to 95 while reducing the margin of error to only 5 n Round up to the nearest integer
To test Ho 100 versus H 100 a simple random sample size of n 24 is obtained from a population that is known to be normally distributed Answer parts a d Click here to view the t Distribution Area in Right Tail a If x 105 3 and s 8 8 compute the test statistic t 2 951 Round to three decimal places as needed b If the researcher decides to test this hypothesis at the x 0 01 level of significance determine the critical values The critical values are 2 807 2 807 Use a comma to separate answers as needed Round to three decimal places as needed c Draw a t distribution that depicts the critical region s Which of the following graphs shows the critical region s in the t distribution O A 1 Q ON OB O C G
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Statistics
To test Ho 100 versus H 100 a simple random sample size of n 24 is obtained from a population that is known to be normally distributed Answer parts a d Click here to view the t Distribution Area in Right Tail a If x 105 3 and s 8 8 compute the test statistic t 2 951 Round to three decimal places as needed b If the researcher decides to test this hypothesis at the x 0 01 level of significance determine the critical values The critical values are 2 807 2 807 Use a comma to separate answers as needed Round to three decimal places as needed c Draw a t distribution that depicts the critical region s Which of the following graphs shows the critical region s in the t distribution O A 1 Q ON OB O C G
To test Ho 20 versus H 20 a simple random sample of size n 16 is obtained from a population that is known to be normally distributed Answer parts a d Click here to view the t Distribution Area in Right Tail RONG to two decimal places as need b Draw a t distribution with the area that represents the P value shaded Which of the following graphs shows the correct shaded region O A B c Approximate the P value Choose the correct range for the P value below OA 0 15 P value 0 20 OB 0 025 P value 0 05 OC 0 10 P value 0 15 0 05 R value 0 10 C Q Q
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Probability
To test Ho 20 versus H 20 a simple random sample of size n 16 is obtained from a population that is known to be normally distributed Answer parts a d Click here to view the t Distribution Area in Right Tail RONG to two decimal places as need b Draw a t distribution with the area that represents the P value shaded Which of the following graphs shows the correct shaded region O A B c Approximate the P value Choose the correct range for the P value below OA 0 15 P value 0 20 OB 0 025 P value 0 05 OC 0 10 P value 0 15 0 05 R value 0 10 C Q Q
To test Ho 60 versus H 60 a random sample of size n 22 is obtained from a population that is known to be normally distributed Complete parts a through d below Click here to view the t Distribution Area in Right Tail to 0 983 Round to three decimal places as needed b If the researcher decides to test this hypothesis at the x 0 05 level of significance determine the critical value s Although technology or a t distribution table can be used to find the critical value in this problem use the t distribution table given Critical Value 1 721 Round to three decimal places Use a comma to separate answers as needed c Draw a t distribution that depicts the critical region Choose the correct answer below OA O B O C ta 0 ta 2 0 ta 2 0
Statistics
Statistics
To test Ho 60 versus H 60 a random sample of size n 22 is obtained from a population that is known to be normally distributed Complete parts a through d below Click here to view the t Distribution Area in Right Tail to 0 983 Round to three decimal places as needed b If the researcher decides to test this hypothesis at the x 0 05 level of significance determine the critical value s Although technology or a t distribution table can be used to find the critical value in this problem use the t distribution table given Critical Value 1 721 Round to three decimal places Use a comma to separate answers as needed c Draw a t distribution that depicts the critical region Choose the correct answer below OA O B O C ta 0 ta 2 0 ta 2 0
To test Ho 107 versus H 107 a simple random sample of size n 35 is obtained Complete parts a through e below Click here to view the t Distribution Area in Right Tail OC 0 01 P value 0 02 OD 0 005 P value 0 01 Interpret the P value Choose the correct answer below A If 1000 random samples of size n 35 are obtained about 4 samples are expected to result in a mean as extreme or more extreme than the one observed if 107 GOLE OB If 100 random samples of size n 35 are obtained about 4 samples are expected to result in a mean as extreme or more extreme than the one observed if 107 OC If 1000 random samples of size n 35 are obtained about 10 samples are expected to result in a mean as extreme or more extreme than the one observed if 107 OD If 1000 random samples of size n 35 are obtained about 4 samples are expected to result in a mean as extreme or more extreme than the one observed if 104 0 e If the researcher decides to test this hypothesis at the x 0 01 level of significance will the researcher reject the null hypothesis Yes No
Statistics
Statistics
To test Ho 107 versus H 107 a simple random sample of size n 35 is obtained Complete parts a through e below Click here to view the t Distribution Area in Right Tail OC 0 01 P value 0 02 OD 0 005 P value 0 01 Interpret the P value Choose the correct answer below A If 1000 random samples of size n 35 are obtained about 4 samples are expected to result in a mean as extreme or more extreme than the one observed if 107 GOLE OB If 100 random samples of size n 35 are obtained about 4 samples are expected to result in a mean as extreme or more extreme than the one observed if 107 OC If 1000 random samples of size n 35 are obtained about 10 samples are expected to result in a mean as extreme or more extreme than the one observed if 107 OD If 1000 random samples of size n 35 are obtained about 4 samples are expected to result in a mean as extreme or more extreme than the one observed if 104 0 e If the researcher decides to test this hypothesis at the x 0 01 level of significance will the researcher reject the null hypothesis Yes No
Several years ago the mean height of women 20 years of age or older was 63 7 inches Suppose that a random sample of 45 women who are 20 years of age or older today results in a mean height of 63 9 inches a State the appropriate null and alternative hypotheses to assess whether women are taller today b Suppose the P value for this test is 0 03 Explain what this value represents c Write a conclusion for this hypothesis test assuming an x 0 10 level of significance CALLY mean marginis 03 7 RIGHTS B There is a 0 03 probability of obtaining a sample mean height of 63 7 inches or taller from a population whose mean height is 63 9 inches OC There is a 0 03 probability of obtaining a sample mean height of exactly 63 9 inches from a population whose mean height is 63 7 inches D There is a 0 03 probability of obtaining a sample mean height of 63 9 inches or shorter from a population whose mean height is 63 7 inches c Write a conclusion for this hypothesis test assuming an x 0 10 level of significance O A Reject the null hypothesis There is not sufficient evidence to conclude that the mean height of women 20 years of age or older is greater today OB Reject the null hypothesis There is sufficient evidence to conclude that the mean height of women 20 years of age or older is greater today O C Do not reject the null hypothesis There is sufficient evidence to conclude that the mean height of women 20 years of age or older is greater today OD Do not reject the null hypothesis There is not sufficient evidence to conclude that the mean height of women 20 years of age or older is greater today
Statistics
Statistics
Several years ago the mean height of women 20 years of age or older was 63 7 inches Suppose that a random sample of 45 women who are 20 years of age or older today results in a mean height of 63 9 inches a State the appropriate null and alternative hypotheses to assess whether women are taller today b Suppose the P value for this test is 0 03 Explain what this value represents c Write a conclusion for this hypothesis test assuming an x 0 10 level of significance CALLY mean marginis 03 7 RIGHTS B There is a 0 03 probability of obtaining a sample mean height of 63 7 inches or taller from a population whose mean height is 63 9 inches OC There is a 0 03 probability of obtaining a sample mean height of exactly 63 9 inches from a population whose mean height is 63 7 inches D There is a 0 03 probability of obtaining a sample mean height of 63 9 inches or shorter from a population whose mean height is 63 7 inches c Write a conclusion for this hypothesis test assuming an x 0 10 level of significance O A Reject the null hypothesis There is not sufficient evidence to conclude that the mean height of women 20 years of age or older is greater today OB Reject the null hypothesis There is sufficient evidence to conclude that the mean height of women 20 years of age or older is greater today O C Do not reject the null hypothesis There is sufficient evidence to conclude that the mean height of women 20 years of age or older is greater today OD Do not reject the null hypothesis There is not sufficient evidence to conclude that the mean height of women 20 years of age or older is greater today
We consider the differences between the reading and writing scores of a random sample of 200 students who took the High School and Beyond Survey The mean and standard deviation of the differences Xread write 0 543 and 8 878 points a Calculate a 95 confidence interval for the average difference between the reading and writing scores of all students Round your answers to two decimal places 1 77 0 69 b Interpret this interval in context Round your answers to two decimal places We are 95 confident that on the reading test students score on average points lower to 0 7 c Does the confidence interval provide convincing evidence that there is a real difference in the average scores Explain Since 0 is included in the interval the confidence interval does not points higher than they do on the writing test provide convincing evidence that there is a real difference in the average scores
Statistics
Statistics
We consider the differences between the reading and writing scores of a random sample of 200 students who took the High School and Beyond Survey The mean and standard deviation of the differences Xread write 0 543 and 8 878 points a Calculate a 95 confidence interval for the average difference between the reading and writing scores of all students Round your answers to two decimal places 1 77 0 69 b Interpret this interval in context Round your answers to two decimal places We are 95 confident that on the reading test students score on average points lower to 0 7 c Does the confidence interval provide convincing evidence that there is a real difference in the average scores Explain Since 0 is included in the interval the confidence interval does not points higher than they do on the writing test provide convincing evidence that there is a real difference in the average scores
The last four years of returns for a stock are as follows Year 1 3 5 Year 2 a What is the average annual return The average return is 28 4 Year 3 12 4 b What is the variance of the stock s returns The variance of the returns is Note Notice that the average return and standard deviation must be entered in percentage format The variance must be entered in decimal format Year 4 3 5 Round to two decimal places c What is the standard deviation of the stock s returns The standard deviation is Round to five decimal places Round to two decimal places
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Statistics
The last four years of returns for a stock are as follows Year 1 3 5 Year 2 a What is the average annual return The average return is 28 4 Year 3 12 4 b What is the variance of the stock s returns The variance of the returns is Note Notice that the average return and standard deviation must be entered in percentage format The variance must be entered in decimal format Year 4 3 5 Round to two decimal places c What is the standard deviation of the stock s returns The standard deviation is Round to five decimal places Round to two decimal places