Week 6 Knowledge

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American Public University *

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302

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Statistics

Date

Feb 20, 2024

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Uploaded by GrandKnowledge13099

Week 6 Knowledge Check Homework Practice Questio... X Attempt 1 of 4 Written Oct 7, 2023 5:48 PM - Oct 7, 2023 10:48 PM Attempt Score 0/20-0% Overall Grade (Highest Attempt) 19/20-95% 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 a = 0.01. The hypotheses are: H0:HD=0 Hi:pp#0 t-Test: Paired Two Sample for Means Placebo Nicotine Mean 16.75| 10.3125 Variance 64.4666/33.29583 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? = > Reject Hy ( > Accept Hy ‘} Accept H, > Do not reject Hy \> Reject Hy Question 2 0/ 1 point An adviser 1s 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 a = 0.05. For the context of this problem, pp=p,.w—Ho1q 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. HO: Uup = 0 HII uD<O

You obtain the following paired sample of 19 students that took the placement test before and after the learning module: New Choose the correct decision and summary and state the p-value. 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

New | o1d LM LM 57 43.1 66.1 [12.8 38.1 [42.2 024 [51.3 > Reject Hy), there 1s enough evidence to support the claim that on average the new online learning module increased placement scores, the p-value = 0.4533. > Reject Hy, there 1s not enough evidence to support the claim that on average the new online learning module increased placement scores, the p-value = 0.2266. = > Do not reject Hy, there 1s not enough evidence to support the claim that on average the new online learning module increased placement scores, the p-value = 0.2266 > Do not reject Hy, there 1s not enough evidence to support the claim that on average the new online learning module increased placement scores, the p-value = 0.4533. Question 3 0/ 1 point Two competing toothpaste brands both claim to produce the best toothpaste for whitening. A dentist randomly samples 48 patients that use Brand A (Group 1) and finds 30 of them are satisfied with the whitening results of the toothpaste. She then randomly samples 45 patients that use Brand B (Group 2) and finds 33 of them are satisfied with the whitening results of the toothpaste. Construct a 99% confidence interval for the difference in proportions and use it to decide if there 1s a significant difference in the satisfaction level of patients. Enter the confidence interval - round to 3 decimal places. ® (-0.356,-.356) <pq-py< ® (0.139,.139) w Hide question 3 feedback Z-Critical Value =NORM.S.INV(.995) = 2.575

LL = (.625-.7333) - 2.575* 625 * .375 N 1333 % .2667 48 45 UL =(.625-.7333) + 2.575* 625 * .375 N 1333 % .2667 48 45 Question 4 0/ 1 point The manager at a sports radio station will be covering several football games over the weekend. She knows that most of her listeners are at least 22 years old and wants to know what age group she should gear her advertisements to for the games. She takes a random sample of 45 listeners from age 22-39 and 55 listeners who are 40 and older and asks them if they are likely to tune in to football games. The "yes" responses are recorded below. Age 22-39 (Group 1) Age 40+ (Group 2) Responded "Yes" to Football 32 44 Sample Size n 45 55 At the 0.05 level of significance, we are attempting to investigate if there 1s a significant difference in the proportion of listeners based on age. Enter the P-Value - round to 4 decimal places. p-value = ® (0.3005, .3005) w Hide question 4 feedback This is a two tailed test because you want to find significant difference. Z=

71111 — .80 \/-T6 % .24 % (1/45 + 1/55) z=-1.03543 Use NORM.S.DIST(-1.03543,TRUE) to find the for the lower tailed test. This value is smaller of the two, thus you will multiply it by 2 for a two tailed test 0.150233*2, this is the p-value you want to use for the conclusion. Question 5 0/ 1 point In a 2-sample z-test for two proportions, you find the following: X1=24n;=200 Xy =17 ny =150 You decide to run a test for which the alternative hypothesis is Hy: p; > p,. Find the appropriate test statistic for the test. Enter the test statistic - round to 4 decimal places. Z: ¥ (0.1919,.1919) ¥ Hide question 5 feedback pl =24/200 p2=17/150 p = (24+17)/(200+150) Z=

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