homework for exm 2.xlsx - question ages 18 texts per day...

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Unformatted text preview: question ages 18 texts per day 48.8 30 what would be the mean of the sampling distribution of the samp 2 μx = μ μ= σ= n= and σx = [ σ / sqrt(n) ] * sqrt[ (N - n ) / (N - 1) ]= 49.6 24.4 92 3 what would be the standard deviation of the sampling distr σx = σ / sqrt(n).= 58.3 μ= 25.3 σ= 111 n= 4 Select the correct words or phrases from the drop-down Suppose you imagine repeatedly taking random samples of size mean μ = 50 and a standard deviation σ = 20. Then the centra the sampling distribution of the sample mean has approx with a mean equal to 50 and a standard deviation equal to 2 5 μ= σ= n= μ= σ= n= What is the probability that the this sample, Ybar , would be 26 2.71260705 0.9966 0.0034 *z score for 2.71 is .9966 i �_¯= �_¯= greater than less than greater than 6 22.3 6.1 20 22.3 1.36400147 �_¯= �_¯= ¯ 100 10 15 100 2.5819889 99.66% 0.34% Find the probability that, for an of 16 candidates, the sample be 105 or more, to four decim P( >105) z < D44 1.93649167 0.9772 0.0228 ¯ P( <105) z > D44 7 *z score for 1.93 is 0.9731 0.9731 n= % **Answer for less than 105 Suppose we were to gather a ra observations from a population a 80% confidence interval for the m the population standard deviatio value from the Student's t distrib three decimal places. 15 80 0.8 0.2 degree of freedom= t-value= 14 1.345030374455 Values from the standard normal (Z) distribution are used most often in practice to form confidence intervals for means because even though the population mean, µ, is unknown, the population standard deviation, σ, is usually known. 8 X **Because if the population std dev is known, w distibution to conduct the confidence interval fo 1 0 Values from the standard normal (Z) distribution are used most often in practice to form confidence intervals for means because even though the population mean, µ, is unknown, the population standard deviation, σ, is usually known. ** 1 ¯= 10 2.28% **Answer for more than 105 1.936491673104 alpha= 9 *z score for 2 is 0.9772 in n= sample sd 15 50.5 20 ****NOT MINUS 1 5.163977794943 % alpha= degree of freedom= t-value= 14 2.144786687918 50.50 plus minus 11 95 0.95 0.05 ¯= n= 11.08 61.5756308313 39.4243691687 15 75 5 sample sd ****NOT MINUS 1 1.290994448736 % alpha= degree of freedom= t-value= 14 2.144786687918 75.00 upper lower 12 n= sample sd 95 0.95 0.05 ¯= 2.77 77.76890770782 72.23109229218 25 118.3 29.4 ****NOT MINUS 1 5.88 % alpha= degree of freedom= t-value= 24 1.710882079909 118.30 upper lower 13 90 0.9 0.1 ¯= n= 10.06 128.3599866299 108.2400133701 51 20.1 3.5 sample sd ****NOT MINUS 1 0.49009802941 % alpha= degree of freedom= t-value= 50 2.677793270941 20.10 upper lower 14 n= sample sd 99 0.99 0.01 ¯= 1.31 21.41238120526 18.78761879475 51 24.1 4.6 ****NOT MINUS 1 0.644128838653 % alpha= degree of freedom= t-value= ¯= upper 50 1.675905025163 24.10 lower 15 26 36 17 12 15 28 17 17 n= sample sd ****NOT MINUS 1 2.878 alpha= degree of freedom= t-value= ¯= upper lower 26.97 21.98 18.72 16.16 19.02 17.87 13.08 14.72 1.08 25.17949875755 23.02050124245 8 21.000 8.142 % 16 90 0.9 0.1 n= sample sd 95 0.95 0.05 7 2.365 21.000 6.807 27.81 14.19 8 18.565 4.370 ****NOT MINUS 1 1.545 % alpha= degree of freedom= t-value= 98 0.98 0.02 7 2.998 ¯= 18.565 upper lower 18 n= 4.632 23.20 13.93 27 43.9 25.3 sample sd ****NOT MINUS 1 4.868987270166 % alpha= degree of freedom= t-value= 26 2.055529438643 43.90 upper lower 95 0.95 0.05 10.01 53.9083466702 33.8916533298 Which of the following is the correct specification of the null and alte H0: Mu < or equal 50, H1: Mu > 50 23 The researcher wishes to test whether the mean amount of televisio Which of the following is the correct specification of the alternativ H1 : μ > 50 What would be a Type I error, in the context of this problem? 24 point estimate of the difference 10.8 Significance level 0.01 #N/A greater than hypothisized mean 42 #N/A sample size 30 #N/A sample mean 52.8 #N/A sample standard de 8.2 #N/A test statistic 7.213906854946 =(D272-D270)TEST STATISTIC= degrees of freedom 29 =D271-1 p value 3.04255147E-08 =T.DIST.RT(D upper tail test=T.DIST.RT() hypothesis decisionReject =IF(D276<D269,"Reject","Do Not Reject") 25 point estimate of the difference 2.2 Significance level 0.05 #N/A greater than hypothisized mean 50 #N/A sample size 30 #N/A sample mean 52.2 #N/A sample standard de 9.7 #N/A test statistic 1.242257346919 =(D287-D285)TEST STATISTIC= degrees of freedom 29 =D286-1 p value 0.112046967185 =T.DIST.RT(D upper tail test=T.DIST.RT() hypothesis decisionDo Not Reject =IF(D291<D284,"Reject","Do Not Reject") 26 27 point estimate of the difference 0.7 less than 28 Significance level 0.01 #N/A hypothisized mean 79.7 #N/A sample size 12 #N/A sample mean 80.4 #N/A sample standard de 0.49 #N/A test statistic 4.948716593054 =(D320-D318)TEST STATISTIC= degrees of freedom 11 =D319-1 p value 0.000218258635 =T.DIST.RT(D upper tail test=T.DIST.RT() hypothesis decisionReject =IF(D324<D317,"Reject","Do Not Reject") point estimate of the difference -0.8 less than Significance level 0.05 #N/A hypothisized mean 79.6 #N/A sample size 66 #N/A sample mean 78.8 #N/A sample standard de 0.39 #N/A test statistic -16.6646941634 =(D335-D333)TEST STATISTIC= degrees of freedom 65 =D334-1 p value 1 =T.DIST.RT(D upper tail test=T.DIST.RT() hypothesis decisionDo Not Reject =IF(D339<D332,"Reject","Do Not Reject") t test= -1.997 1.997 z test -1.96 1.96 29 13.99 20.14 22.86 21.88 19.64 29.96 16.52 27.24 point estimate of the difference 1.529 equal to 31 Significance level 0.05 #N/A hypothisized mean 20 #N/A sample size 8 =COUNT(A349:A356) sample mean 21.529 =AVERAGE(A349:A356) sample standard de 5.251 =STDEV.S(A349:A356) test statistic 0.82 =(D352-D350)TEST STATISTIC= degrees of freedom 7 =D351-1 p value 0.219 =T.DIST.RT(D upper tail test=T.DIST.RT() hypothesis decisionDo Not Reject =IF(D356<D349,"Reject","Do Not Reject") t test= -1.997 1.997 z test -1.96 1.96 point estimate of the difference 1323.4 less than Significance level 0.05 #N/A hypothisized mean 79.6 #N/A sample size 188 #N/A sample mean 1403 #N/A sample standard de 480 #N/A test statistic 37.80322207571 =(D369-D367)TEST STATISTIC= degrees of freedom 187 =D368-1 p value 8.25342775E-90 =T.DIST.RT(D upper tail test=T.DIST.RT() hypothesis decisionReject =IF(D373<D366,"Reject","Do Not Reject") z test for 95% 1.96 largest value 1471.61 xts per day std dev sample size 61 24.3 ling distribution of the sample mean, to one decimal place (N - n ) / (N - 1) ]= 49.6 on of the sampling distribution of the sample mean, to one decimal place 2.40136973 s from the drop-down boxes to complete the following sentence. g random samples of size 100 from a population with a σ = 20. Then the central limit theorem says that mple mean has approximately a normal distribution rd deviation equal to 2 hat is the probability that the average ACT score of is sample, Ybar , would be less than 24? z score for 2.71 is .9966 in distro table nd the probability that, for any given random sample 16 candidates, the sample mean score, Ybar would e 105 or more, to four decimal places. z score for 2 is 0.9772 in distro table Answer for more than 105 z score for 1.93 is 0.9731 in distro tabl Answer for less than 105 uppose we were to gather a random sample of 15 bservations from a population and wished to calculate an 0% confidence interval for the mean, µ, in the case where e population standard deviation, σ, is unknown. Enter the lue from the Student's t distribution that we would use, to ree decimal places. ion are used most for means because nown, the nown. e population std dev is known, we use zonduct the confidence interval for unknown mean ion are used most for means because nown, the nown. 16 An alumni association at a university wants to gather data on the an employed as accountants who earned their CPA licenses one year from university records of all accounting students who graduated a out a survey to the selected individuals in the sample and received alumni association used the responses to calculate a 95% confiden accountants one year after obtaining their CPA licenses, finding the year. Which of the following statements is the correct interpretation carefully! accountants one year after obtaining their CPA licenses, finding the year. Which of the following statements is the correct interpretation carefully! Answer The alumni association can be 95% confident, based on the method the true mean salary of the university's accounting graduates one y is between $64,000 and $89,000 per year." fication of the null and alternative hypotheses? Read carefully! e mean amount of television watched daily by young men is greater than 50 minutes. fication of the alternative hypothesis? ext of this problem? t=(¯−�_0)/ (/√) of the difference ST STATISTIC= per tail test=T.DIST.RT() ,"Reject","Do Not Reject") of the difference t=(¯−�_0)/ (/√) ST STATISTIC= per tail test=T.DIST.RT() ,"Reject","Do Not Reject") of the difference t=(¯−�_0)/ (/√) ST STATISTIC= per tail test=T.DIST.RT() ,"Reject","Do Not Reject") of the difference t=(¯−�_0)/ (/√) ST STATISTIC= per tail test=T.DIST.RT() ,"Reject","Do Not Reject") Common Confidence interval with associated z score 90% 1.645 95% 1.96 99% 2.576 of the difference ST STATISTIC= t=(¯−�_0)/ (/√) per tail test=T.DIST.RT() ,"Reject","Do Not Reject") of the difference ST STATISTIC= t=(¯−�_0)/ (/√) per tail test=T.DIST.RT() ,"Reject","Do Not Reject") 1 A _Sampling distribution_is the distribution of a calcul statistic, such as a sample mean or sample proportion, w consider taking repeated samples of a given size, n, from same population or process and calculating that statistic time. 6 μ= σ= n= 68 2.8 �_¯= 16 �_¯= 68 0.7 ¯ P( >69.2) 1.71428571 z < D44 0.9563 ****** 0.0437 4.37% ¯ P( <105) z > D44 7 52.8571428571 0.9731 n= % 19 99 0.99 0.01 alpha= degree of freedom= t-value= 9 18 2.878 A confidence interval is a range of values within whic confident (based on the method used to calculate the that the actual value of a population parameter (such population mean) will fall. 1 ** to gather data on the annual earnings of former students CPA licenses one year ago. After taking a random sample dents who graduated a year ago, the alumni association sent e sample and received 75 responses. An analyst for the alculate a 95% confidence interval for the mean salary of PA licenses, finding the interval to be $64,000 to $89,000 per he correct interpretation of this interval in context? Read nt, based on the method used to calculate the interval, that ounting graduates one year after obtaining their CPA licenses 50 minutes. ssociated z score stribution of a calculated sample proportion, when we a given size, n, from the ulating that statistic each What would be the probability of seeing a sample mean height at least this large if the population mean height were really 68 inches? Report your answer to four decimal places. *z score for 2 is 0.9772 in distro table **Answer for more than 105 *z score for 1.93 is 0.9731 in distro tabl **Answer for less than 105 Suppose we were to gather a random sample of 19 observations from a population and wished to calculate an 99% confidence interval for the mean, µ, in the case where the population standard deviation, σ, is unknown. Enter the value from the Student's t distribution that we would use, to three decimal places. e of values within which we are d used to calculate the interval) ation parameter (such as μ, the question 1 In a recent survey, out of a random sample of 1,317 adults 18 and over in the United States, 596 reported that acquiring wealth was "of more than average importance" in their lives. Calculate a 99% confidence interval for the true proportion of adults in the United States who b acquiring wealth is "of more than average importance," to three decimal places. Take all calcula the answer to four (4) decimal places. ̂=/ 2 Proportion 0.4525 x n 596 1317 z-critical value 99%= C.I. lower C.I. upper 2.576 Common Confidence interval with associated 90% 1.645 95% 1.96 99% 2.576 0.417 0.488 Management of a company that operates a nationwide network of pharmacies is interested in o vaccine (the "flu shot") in its stores. Before the company invests in the necessary supplies and t wants to estimate the size of its potential market. After collecting data from a survey of 200 ran selected customers, the company found that 96 of them reported that they get the flu shot eve some other provider, e.g., a doctor or another pharmacy). The company used this information t 95% confidence interval, obtaining the endpoints 0.411 and 0.549. ̂=/ Proportion 0.4800 x n 96 200 z-critical value 95%= C.I. lower 1.96 0.411 48% Common Confidence interval with associated 90% 1.645 95% 1.96 99% 2.576 Answer The company can be 95% confident, based on the m to calculate the interval, that the true proportion o 41.08% who get the flu shot every year is between 0.411 an is, between 41.1% and 54.9% all customers get the year. C.I. upper 3 n a recent survey, 472 CEOs of medium and large companies were asked whether they possesse degree. In the sample, 223 had MBA degrees. Give the lower bound of the 99% confidence inte true proportion of all CEOs who have MBAs. Take all intermediate calculations to three decimal ̂=/ x n Proportion 0.472 47% 223 472 z-critical value 99%= C.I. lower C.I. upper 4 0.549 The company can be 95% confident, based on the m to calculate the interval, that the true proportion o who get the flu shot every year is between 0.411 an 54.92% is, between 41.1% and 54.9% all customers get the year. 2.576 0.413 0.532 Common Confidence interval with associated 90% 1.645 95% 1.96 99% 2.576 Answer C.I. lower0.413 41.33% 53.17% A certain transportation system of buses and commuter trains is heavily utilized so that it is not check every traveler's ticket. Rather, only a small, randomly selected group of travelers on any g be asked to show their tickets. Suppose that in a random sample of 671 train travelers is selecte them admitted they did not buy a ticket. Find the upper bound of a 95% confidence interval for proportion of all train travelers who do not buy tickets. Take all calculations to three decimal pla your answer to three decimal places. ̂=/ Proportion 0.103 10% Common Confidence interval with associated 90% 1.645 ̂=/ 5 x n 69 671 z-critical value 95%= C.I. lower C.I. upper 1.96 0.080 0.126 95% 99% 1.96 2.576 Answer C.I. upper0.126 7.98% 12.58% In commercials for oral hygiene products such as toothpaste, mouthwash, and whitening strips, sometimes hear claims like "over 9 out of 10 dentists recommend (whatever is being advertised 10? Why does it seem like there is always one dentist who doesn't like the product?The answer advertiser who made the commercial most likely doesn't mean that, literally, 10 dentists were a dentist indicated that he or she would not recommend the product. What the advertiser is reall making a claim about a population proportion.How could the advertiser make this claim using m research? One way is to conduct a hypothesis test to determine if the true proportion of dentist recommend the product is greater than 0.90.Suppose that the advertiser surveys a random sam dentists and found that 91 of them indicated that they would recommend a particular product. test statistic and state the conclusion at the 0.05 level (two selections; read carefully): Proportion Common Confidence interval with associated p 0.910 91% 90% 1.645 ̂=/ level of significance 0.05 95% 1.96 x 91 99% 2.576 n 100 Ho:p=.90 0.90 Ha:p>.90 0.90 std dev of p= 0.03 test statistic is z= 0.333 Fail To Reject the null hypothisis at 0.05 level of significance z-critical value 90%= C.I. lower C.I. upper 1.645 0.863 0.957 Answer z= 0.333 Fail To Reject the null hypothisis at 0.05 level o 86.29% The advertiser CANNOT conclude that more than 9 95.71% dentists would recommend the product 6 According to the Centers for Disease Control and Prevention , in 2008, about 11% of all America age and older used antidepressant medication (examples include Prozac, Paxil, Cymbalta, and Z medical researcher wants to know if this proportion has changed since then.A random sample o are randomly selected from this age group and it is determined that 130 of them used some kin antidepressant medication. Conduct a hypothesis test for a proportion at the 0.05 level of signifi the test statistic and the conclusion (two selections). Proportion Common Confidence interval with associated p 0.130 13% 90% 1.645 ̂=/ level of significance 0.05 95% 1.96 x 130 99% 2.576 n 1000 Ho:p=11% 0.11 Ha:p>11% 0.11 std dev of p= 0.009894443 Increased test statistic is z= 2.021 Reject the null hypothisis at 0.05 level of significance z-critical value 95%= C.I. lower C.I. upper 7 ̂=/ 1.96 0.109 0.151 Answer z= 2.021 Reject the null hypothisis at 0.05 level of sign 10.92% usage has increased since 15.08% Many businesses use direct mail (oftentimes called "junk mail") to advertise their products and with the declining volume of physical mail, the United States Postal Service has encouraged its u few years. Direct mail advertising involves sending solicitations (credit card promotions, special dealerships and furniture stores, grocery store circulars, etc.) to many thousands of people in ho proportion of those people will end up buying products or services from them. These businesse consumer information data sets from "data brokers," that, in turn, gather and consolidate consu various sources such as warranty cards, surveys, government and public records, social media si Linkedin and Facebook, and other places. In case you are interested, you can find more informa , although you do not nee answer the question.Suppose a company decides to test out a new layout on a flier before fully a nationwide mailing campaign. It mails the flier to a sample of 1,198 people randomly selected Proportion p 0.108 level of significance 0.05 x 129 11% Common Confidence interval with associated 90% 1.645 95% 1.96 99% 2.576 test statistic is n Ho:p=11% Ha:p>11% std dev of p= z= z-critical value 95%= C.I. lower C.I. upper 8 1198 0.11 0.11 0.009039886 decreased -0.257 Reject the null hypothisis at 0.05 level of significance 1.96 0.090 0.125 Answer C.I. upper0.125 9.01% 12.52% When we are comparing two population means from two groups (suppose they are called Grou 2), which of the following statistics do we use to make a confidence interval or perform a hypot difference in means? Proportion Common Confidence interval with associated p 0.108 11% 90% 1.645 ̂=/ level of significance 0.05 95% 1.96 x 129 99% 2.576 n 1198 Ho:p=11% 0.11 Ha:p>11% 0.11 std dev of p= 0.009039886 decreased test statistic is z= -0.257 Reject the null hypothisis at 0.05 level of significance z-critical value 95%= C.I. lower C.I. upper 1.96 0.090 0.125 Answer A confidence interval or perform a hypothesis test f PARTIAL POINTS 9.01% in means the quantity is. 12.52% �_1−�_2 9 A credit card company wants to determine whether a new incentive could increase average cus spending. The company took two random samples of 500 cardholders each from its database. T calculated a 95% confidence interval to determine if the difference in mean spending between t an incentive (Group 1: "Incentive") and the group without an incentive (Group 2: "No incentive" The interval was $46.79 to $193.06. Which of the following gives the correct interpretation of th Read carefully! Proportion Common Confidence interval with associated ̂=/ p 0.258 26% 90% 1.645 level of significance 0.05 95% 1.96 x 129 99% 2.576 n 500 Ho:p=11% 0.11 Ha:p>11% 0.11 std dev of p= 0.013992855 Increased test statistic is z= 10.577 Reject the null hypothisis at 0.05 level of significance z-critical value 95%= C.I. lower C.I. upper 10 ̂=/ 1.96 0.220 0.296 Answer The company can be 95% confident that the true di mean spending between customers offered an ince 21.96% those not offered an incentive is between $46.79 a 29.64% Because the interval does not include the value of 0 can conclude that the i...
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