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

Before After 55.9 44.7 56.8 50.6 71 63.4 80.6 68.9 59.8 35.5 73.1 77 49.9 38.4 56.2 55.4 64.3 55.6 Find the p-value. Round answer to 4 decimal places. p-value: ___ 0.0038 ___ Hide question 2 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 Before Variable 2 Range: is After 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) 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