Homework 6 Attempt 1

Question 1 0 / 1 point A physician wants to see if there was a difference in the average smokers' daily cigarette consumption after wearing a nicotine patch. The physician set up a study to track daily smoking consumption. In the study, the patients were given a placebo patch that did not contain nicotine for 4 weeks, then a nicotine patch for the following 4 weeks. Test to see if there was a difference in the average smoker's daily cigarette consumption using α = 0.01. The hypotheses are: H 0 : μ D = 0 H 1 : μ D ≠ 0 t-Test: Paired Two Sample for Means Placebo Nicotine Mean 16.75 10.3125 Variance 64.4666 7 33.2958 3 Observations 16 16 Pearson Correlation 0.6105 Hypothesized Mean Difference 0 df 15

t Stat 4.0119 P(T<=t) one-tail 0.0006 t Critical one-tail 2.6025 P(T<=t) two-tail 0.0011 t Critical two-tail 2.9467 What is the correct decision? Accept H 1 Accept H 0 Do not reject H 0 Reject H 1 Reject H 0 Hide question 1 feedback The p-value for a two tailed test is 0.0011. This is given to you in the output. No calculations are needed. 0.0011 < .01, Reject Ho, this is significant. Question 2 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: Choose the correct decision and summary and state the p-value. , there is not enough evidence to support the claim that on average the new online learning module increased placement scores, the p-value = 0.2266 , there is enough evidence to support the claim that on average the new online learning module increased placement scores, the p-value = 0.4533. , there is not enough evidence to support the claim that on average the new online learning module increased placement scores, the p-value = 0.4533. , there is not enough evidence to support the claim that on average the new online learning module increased placement scores, the p-value = 0.2266. 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 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 0.2266 Question 3 1 / 1 point New LM Old LM 58.1 55.8 58.3 53.7 83.6 76.6 49.5 47.5 51.8 48.9 20.6 11.4 35.2 30.6 46.7 54 22.5 21 47.7 58.5 51.5 42.6 76.6 61.2 29.6 26.8 14.5 12.5 43.7 56.3 57 43.1 66.1 72.8

A survey was given to college students nationwide. The question asked was: "Do you completely abstain from alcohol or not?" Of a random sample of 212 students at community colleges (Group 1), 87 claimed to completely abstain. Of a random sample of 390 students at 4-year colleges (Group 2), 103 claimed to completely abstain. At α=0.10 , is there evidence to claim a difference in the two proportions exists? Select the correct alternative hypothesis and decision. H 1 : p 1 ≠ p 2 ; Do not reject the null hypothesis. H 1 : p 1 ≠ p 2 ; Reject the null hypothesis. H 1 : p 1 > p 2 ; Do not reject the null hypothesis. H 1 : p 1 > p 2 ; Reject the null hypothesis. H 1 : p 1 < p 2 ; Do not reject the null hypothesis. H 1 : p 1 < p 2 ; Reject the null hypothesis. Hide question 3 feedback This is a two tailed test because you want to find significant evidence. z = .410377−.264103.315615 ∗ .684385 ∗ (1/212+1/390) z = 3.688439 Use NORM.S.DIST(3.688439,TRUE) to find the for the lower tailed test. 0.999887, then 1 - 0.999887 = 0.000113 Multiply by 2 for a two tailed test 0.000113*2, this is the p-value you want to use for the conclusion. Question 4 1 / 1 point A researcher took a random sample of 200 new mothers in the United States (Group 1) and found that 16% of them experienced some form of postpartum depression. Another random sample of 200 new mothers in France (Group 2), where mothers receive 16

weeks of paid maternity leave, found that 11% experienced some form of postpartum depression. Can it be concluded that the percent of women in the United States experience more postpartum depression than the percent of women in France? Use α=0.05 . Select the correct alternative hypothesis and decision . H 1 : p 1 ≠ p 2 ; Do not reject the null hypothesis. H 1 : p 1 ≠ p 2 ; Reject the null hypothesis. H 1 : p 1 > p 2 ; Do not reject the null hypothesis. H 1 : p 1 > p 2 ; Reject the null hypothesis. H 1 : p 1 < p 2 ; Do not reject the null hypothesis. H 1 : p 1 < p 2 ; Reject the null hypothesis. Hide question 4 feedback This is an upper tailed test because because of the keyword more. z = .16−.11.135 ∗ .865 ∗ (1/200+1/200) z = 1.463171 Use NORM.S.DIST(1.463171,TRUE) to find the for the lower tailed test. 0.92829, then 1 - 0.92829, this is the p-value you want to use for an upper tailed test. Question 5 1 / 1 point A high school is running a campaign against the over-use of technology in teens. The committee running the campaign decides to look at the difference in social media usage between teens and adults. They take a random sample of 200 teens in their city (Group 1) and find that 85% of them use social media, and then take another random sample of 180 adults in their city (Group 2) and find that 55% of them use social media. Find a 90% confidence interval for the difference in proportions. Enter the confidence interval - round to 4 decimal places. ___ 0.2262 ___ (50 %)