NewLMOldLM58.155.858.353.783.676.649.547.551.848.920.611.435.230.646.75422.52147.758.551.542.676.661.229.626.814.512.543.756.35743.166.172.838.142.242.451.3An adviser is testing out a new online learning module for a placement test. They wish to test the claim that on average thenew online learning module increased placement scores at a significance level of α= 0.05. For the context of thisproblem, μ=μ–μwhere the first data set represents the new test scores and the second data set represents old testscores. Assume the population is normally distributed.H : μ= 0H : μ< 0You obtain the following paired sample of 19 students that took the placement test before and after the learning module:Choose the correct decision and summary and state the p-value.Question 1 of 201.0/ 1.0 PointsDnewold0D1DA. Do not reject H, there is not enough evidence to support the claim that on average the new online learning module increasedplacement scores, the p-value = 0.4533.B. Reject H, there is enough evidence to support the claim that on average the new online learning module increasedplacement scores, the p-value = 0.4533.C. Reject H, there is not enough evidence to support the claim that on average the new online learning module increasedplacement scores, the p-value = 0.2266.000

D. Do not reject H, there is not enough evidence to support the claim that on average the new online learning module increasedplacement scores, the p-value = 0.2266Answer Key:D

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 -> OKVariable 1 Range: is New LMVariable 2 Range: is Old LMThe Hypothesized Mean Difference is 0 and make sure you click Labels in the first row and click OK. You will get an output andthis is the p-value you are looking for.P(T<=t) one-tail 0.2266A physician wants to see if there was a difference in the average smokers' daily cigarette consumption after wearing anicotine patch. The physician set up a study to track daily smoking consumption. In the study, the patients were given aplacebo patch that did not contain nicotine for 4 weeks, then a nicotine patch for the following 4 weeks. Test to see if therewas a difference in the average smoker’s daily cigarette consumption using α= 0.01. The hypotheses are:H: μ= 0H1: μ≠t-Test: Paired Two Sample for MeansPlacebo NicotineMean16.7510.3125Variance64.46667 33.29583Observations1616Pearson Correlation0.6105Hypothesized Mean Difference0df15t Stat4.0119P(T<=t) one-tail0.0006t Critical one-tail2.6025P(T<=t) two-tail0.0011t Critical two-tail2.9467What is the correct decision?0