Quiz 6 Attempt 1

Question 1 1 / 1 point An adviser is testing out a new online learning module for a placement test. They wish to test the claim that on average the new online learning module increased placement scores at a significance level of α = 0.05. For the context of this problem, μ D =μ new –μ old where the first data set represents the new test scores and the second data set represents old test scores. Assume the population is normally distributed. H 0 : μ D = 0 H 1 : μ D < 0 You obtain the following paired sample of 19 students that took the placement test before and after the learning module: Find the p-value. Round answer to 4 decimal places. p-value = ___ 0.1987 ___ Hide question 1 feedback Copy and paste the data into Excel. Use the Data Analysis Toolpak in Excel. Data - > Data Analysis -> scroll to where is says t:Test: Paired Two Samples for Means -> OK Variable 1 Range: is New LM Variable 2 Range: is Old LM The Hypothesized Mean Difference is 0 and make sure you click Labels in the first row and click OK. You will get an output and this is the p-value you are looking for. P(T<=t) one-tail 0.1987 Question 2 1 / 1 point New LM Old LM 57.1 55.8 58.3 51.7 83.6 76.6 50.5 47.5 51.5 48.6 20.6 15.5 35.2 29.9 46.7 54 23.5 21 48.8 58.5 53.1 42.6 76.6 61.2 29.6 26.3 14.5 11.4 43.7 56.3 57 46.1 66.1 72.9

A manager wishes to see if the time (in minutes) it takes for their workers to complete a certain task is different if they are wearing earbuds. A random sample of 20 workers' times were collected before and after wearing earbuds. Test the claim that the time to complete the task will be different, i.e. meaning has production differed at all, at a significance level of α = 0.01 For the context of this problem, μ D = μ before −μ after where the first data set represents before earbuds and the second data set represents the after earbuds. Assume the population is normally distributed. The hypotheses are: H 0 : μ D = 0 H 1 : μ D 0 You obtain the following sample data: Before After 69 65.3 69.5 61.6 39.3 21.4 66.7 60.4 38.3 46.9 85.9 76.6 70.3 77.1 59.8 51.3 72.1 69 79 83 61.7 58.8

p-value you are looking for. P(T<=t) two-tail 0.0038 Question 3 1 / 1 point In a 2-sample z-test for two proportions, you find a p -value of 0.0081. In a two-tailed test with α=0.01 , what decision should be made? Not enough information to tell Do not reject the null. Reject the null. Question 4 1 / 1 point While running a 2-sample z-test for two proportions, you find a test statistic of z=1.946 . If the test is two-tailed and α=0.05 , what decision will you make? Reject the null. Do not reject the null. Not enough information to solve Hide question 4 feedback =NORM.S.DIST(1.946,TRUE) p-value for an upper tailed test = 0.97417263 Lower tailed test = 1 - 0.97417263 = 0.02582737 Multiply this by 2 to get the p-value for the conclusion. Question 5 0 / 1 point The makers of a smartphone have received complaints that the face recognition tool often doesn't work, or takes multiple attempts to finally unlock the phone. They've upgraded to a new version and are claiming the tool has improved. To test the claim, a critic takes a random sample of 80 users of the new version (Group 1) and 75 users of the old version (Group 2). He finds that the face recognition tool works on the first try 75% of the time in the new version and 60% of the time in the old version. Can it be concluded that the new version is performing better? Test at α=0.10.

Hypotheses: H 0 : p 1 = p 2 H 1 : p 1 > p 2 In this scenario, what is the test statistic? Round to four decimal places. z =_____ ___1.806421 ___ (1.9964 ) Hide question 5 feedback z = .75−.60.677419 ∗ .322581 ∗ (1/80+1/75) z = 1.9964 Question 6 1 / 1 point A two-tailed test is one where: negative sample means lead to rejection of the null hypothesis results in only one direction can lead to rejection of the null hypothesis results in either of two directions can lead to rejection of the null hypothesis no results lead to the rejection of the null hypothesis Question 7 1 / 1 point Results from previous studies showed 76% of all high school seniors from a certain city plan to attend college after graduation. A random sample of 200 high school seniors

that from the time shoots are planted 90 days on average are required to obtain the first berry. A corporation that is interested in marketing the product tests 60 shoots by planting them and recording the number of days before each plant produces its first berry. The sample mean is 92.3 days and the SD is 6.66 days. The corporation wants to know if the mean number of days is different from the 90 days claimed? Use alpha = .01. Yes, because the p-value = .0097 No, because the p-value = .9873 No, because the p-value =.0048 Yes, because the p-value = .0048 No, because the p-value = .0097 Hide question 9 feedback The hypothesized value is 90 and this is a two tailed test because the keyword is different. T Stat = 92.3−906.6660 T stat = 2.675 Use =T.DIST.2T(2.675,59) to find the p-value. Question 10 1 / 1 point Which if the following values defaults to alpha when no significance level is given? 0.50 0.01 0.10

0.05 Question 11 1 / 1 point A researcher takes sample temperatures in Fahrenheit of 17 days from New York City and 18 days from Phoenix. Test the claim that the mean temperature in New York City is different from the mean temperature in Phoenix. Use a significance level of α=0.10. Assume the populations are approximately normally distributed with unequal variances. You obtain the following two samples of data. New York City Phoenix 98 93.2 95.5 72 92.2 86.8 102 120.1 85.4 114.4 78 93.7 85.4 89.7 75.4 104.7

Reject H 0 . There is evidence that the mean temperature in New York City is different from the mean temperature in Phoenix. Hide question 11 feedback Running the test using the Data Analysis Toolpak the p-value for a two tailed test is 0.0583. This is less than .10. Running the test using the Data Analysis Toolpak the p-value for a two tailed test is 0.0583. This is less than .10. Question 12 1 / 1 point A pet store owner believes that dog owners spend more on average than cat owners on their pets. The owner randomly records the sales of 40 customers that said they only owned dogs and 40 customers that said they only owned cats. What is the correct decision and summary based on the Excel output below? Assume the populations are approximately normally distributed with unequal variances. t-Test: Two-Sample Assuming Unequal Variances Dog Cat Mean 57.1695 53.82225 Variance 588.088 573.7205 Observations 40 40 Hypothesized Mean Difference 0 df 78 t Stat 0.6211 P(T<=t) one-tail 0.2682

t Critical one-tail 1.6646 P(T<=t) two-tail 0.5363 t Critical two-tail 1.9908 What is the correct test statistic? 1.6646 0.6211 0.2682 1.9908 0.5860 Hide question 12 feedback No Calculations are needed. Looking at the output the t Stat is 0.6211 Question 13 1 / 1 point Two random samples are taken from private and public universities (out-of-state tuition) around the nation. The yearly tuition is recorded from each sample and the results can be found below. Test to see if the mean out-of-state tuition for private institutions is statistically significantly higher than public institutions. Assume unequal variances. Use a 1% level of significance. Private Institutions (Group 1 ) Public Institutions (Group 2) 43,120 25,469 28,190 19,450 34,490 18,347

What are the correct hypotheses for this problem? H 0 : μ 1 ≥ μ 2 ; H 1 : μ 1 ≤ μ 2 H 0 : μ 1 = μ 2 ; H 1 : μ 1 > μ 2 H 0 : μ 1 ≤ μ 2 ; H 1 : μ 1 ≥ μ 2 H 0 : μ 1 = μ 2 ; H 1 : μ 1 ≠ μ 2 H 0 : μ 1 < μ 2 ; H 1 : μ 1 = μ 2 H 0 : μ 1 ≠ μ 2 ; H 1 : μ 1 = μ 2 Hide question 13 feedback Upper tailed test because of the keyword higher Upper tailed test because of the keyword higher Question 14 1 / 1 point In a random sample of 50 Americans five years ago (Group 1), the average credit card debt was $5,598. In a random sample of 50 Americans in the present day (Group 2), the average credit card debt is $6,499. Let the population standard deviation be $1,155 five years ago, and let the current population standard deviation be $1,699. Using a 0.01 level of significance, test if there is a difference in credit card debt today versus five years ago. What is the test statistic? (Round to 4 decimal places) z =___ ___ Answer: -3.1011 Hide question 14 feedback z -stat = 5598−64991155250+1699250

Question 15 1 / 1 point The CEO of a large manufacturing company is curious if there is a difference in productivity level of her warehouse employees based on the region of the country the warehouse is located. She randomly selects 35 employees who work in warehouses on the East Coast (Group 1) and 35 employees who work in warehouses in the Midwest (Group 2) and records the number of parts shipped out from each for a week. She finds that East Coast group ships an average of 1287 parts and knows the population standard deviation to be 348. The Midwest group ships an average of 1449 parts and knows the population standard deviation to be 298. Using a 0.01 level of significance, test if there is a difference in productivity level. What are the correct hypotheses for this problem? H0: μ1= μ2 ; H1: μ1 ≠ μ2 H0: μ1 ≤ μ2 ; H1: μ1 ≥ μ2 H0: μ1 ≠ μ2 ; H1: μ1 = μ2 H0: μ1 ≥ μ2 ; H1: μ1 ≤ μ2 H0: μ1 = μ2 ; H1: μ1 = μ2 H0: μ1 > μ2 ; H1: μ1 = μ2 Hide question 15 feedback This is a two tailed test because of the keyword difference. Question 16 1 / 1 point A professor wants to know whether or not there is a difference in comprehension of a lab assignment among students depending on if the instructions are given all in text, or if they are given primarily with visual illustrations. She randomly divides her class into two groups of 15, gives one group instructions in text and the second group instructions with visual illustrations. The following data summarizes the scores the students received on a test given after the lab. Let the populations be normally distributed with a population standard deviation of 5.32 points for both the text and visual illustrations.

Answer: 0.0212 Hide question 16 feedback Copy and paste the data into Excel. Use the Data Analysis Toolpak in Excel. Data - > Data Analysis -> scroll to where is says z:Test 2 Samples for Means -> OK Variable 1 Range: highlight Text Variable 2 Range: highlight Visual Illustrations The Hypothesized Mean Difference is 0. Variable 1 Variance: You are given the population SD you need to take 5.32 2 and use this for the population Variance for Variable 1 and 2. Make sure you click Labels in the first row and click OK. You will get an output and this is the p-value you are looking for. 0.0212 Question 17 1 / 1 point In a random sample of 29 residents living in major cities on the West Coast (Group 1) and 29 residents living in major cities on the East Coast (Group 2), each individual was asked their age. The results can be seen in the table below. The population standard deviation of the age in West Coast cities is known to be 10.95 years and in East Coast cities is known to be 9.67 years. Assume the populations are normally distributed. Run a test at a 0.05 level of significance to test if west coast cities are, on average, older. West Coast (Group 1) East Coast (Group 2) 25 35 47 45 18 37

38 20 30 19 52 26 52 79 61 46 43 29 22 55 34 25 35 35 55 36 60 53 68 41 20 50 34 32 36 38 37 26 42 44 60 19 71 28 54 27

2.8855 Question 18 1 / 1 point According to the February 2008 Federal Trade Commission report on consumer fraud and identity theft, 23% of all complaints in 2007 were for identity theft. In that year, Alaska had 321 complaints of identity theft out of 1,432 consumer complaints. Does this data provide enough evidence to show that Alaska had a lower proportion of identity theft than 23%? The hypotheses are: H 0 : p = 23% H 1 : p < 23% What is a type I error in the context of this problem? Reference: Federal Trade Commission, (2008). Consumer fraud and identity theft complaint data: January-December 2007. Retrieved from website: http://www.ftc.gov/opa/2008/02/fraud.pdf. It is believed that more than 23% of Alaskans had identity theft and there really were 23% or more that experience identity theft. It is believed that less than 23% of Alaskans had identity theft and there really were 23% or less that experienced identity theft. It is believed that less than 23% of Alaskans had identity theft even though there really were 23% or more that experienced identity theft. It is believed that more than 23% of Alaskans had identity theft even though there really were less than 23% that experienced identity theft.

Question 19 1 / 1 point Match the following symbol with the correct phrase. α significance level confidence level parameter power P(Type II Error) Hide question 19 feedback If you have a 95% confidence level then alpha = 1 .95 = .05. .05 is the significance level for alpha. Question 20 0 / 1 point The workweek for adults in the US that work full time is normally distributed with a mean of 47 hours. A newly hired engineer at a start-up company believes that employees at start-up companies work more on average then working adults in the US. She asks 15 engineering friends at start-ups for the lengths in hours of their workweek. Their responses are shown in the table below. Test the claim using a 5% level of significance. See Excel for Data. Work Hour Data The hypotheses for this problem are: H 0 : μ = 47 H 1: μ > 47