# MGMT 650 Quiz 8

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Quiz 8 Question 1 The director for Weight Watchers International wants to determine if the changes in their program results in better weight loss. She selected 25 Weight Watcher members at random and compared their weight 6 months later to weight at the start of the program. The results are in this excel file: Weight Watchers.xlsx (The weight in the column labeled "After" represents their weights six months later and "Before" represents their weight at the start of the six-month period.) The director used .05 as the significance level. Use Excel to test. For each paired difference, compute After - Before. In Data Analysis, t-Test: Paired Two Sample for means, select the After data for Variable 1 Range. Note that the critical value output by Data Analysis for this test is always positive. In this problem, the sign of the critical value is negative corresponding to a 1-tailed test with lower reject region and negative lower critical value. What are the Null and Alternate Hypotheses? (Keep the Excel file open for the next parts of the question.) Question options: H0: After - Before ≥ 0; HA: After - Before > 0 H0: After - Before ≥ 0; HA: After - Before < 0 H0: After - Before = 0; HA: After - Before ≠ 0 H0: After - Before > 0; HA: After - Before ≤ 0 Question 2 Refer to the Weight Watchers file here: Weight Watchers.xlsx Use Excel to test. For each paired difference, compute After - Before. In Data Analysis, t-Test: Paired Two Sample for means, select the After data for
Variable 1 Range. Note that the critical value output by Data Analysis for this test is always positive. In this problem, the sign of the critical value is negative corresponding to a 1-tailed test with lower reject region and negative lower critical value. What is the t-statistic (t-score)? Question options: -9.321 -7.087 7.087 9.321 Question 3 Refer to the Weight Watchers file here: Weight Watchers.xlsx Use Excel to test. For each paired difference, compute After - Before. In Data Analysis, t-Test: Paired Two Sample for means, select the After data for Variable 1 Range. Note that the critical value output by Data Analysis for this test is always positive. In this problem, the sign of the critical value is negative corresponding to a 1-tailed test with lower reject region and negative lower critical value. What is the critical t-value? Question options: 1.711 or -1.711 7.861 or -7.861 5.614 or -5.614 3.645 or -3.645 Question 4 Refer to the Weight Watchers file here: Weight Watchers.xlsx
Use Excel to test. For each paired difference, compute After - Before. In Data Analysis, t-Test: Paired Two Sample for means, select the After data for Variable 1 Range. Note that the critical value output by Data Analysis for this test is always positive. In this problem, the sign of the critical value is negative corresponding to a 1-tailed test with lower reject region and negative lower critical value. What is your conclusion? Question options: Do not reject the NULL Hypothesis because the actual value is less than the critical value Reject the NULL Hypothesis because the actual value is less than the critical value Do not reject the NULL Hypothesis because the actual value is greater than critical value Reject the NULL Hypothesis because the actual value is greater than critical value Question 5 A new drug is created to help people with depression. When testing the drug, what are the null and alternative hypotheses? Question options: H0: As a result of 300mg./day of the ABC drug, there will be no significant difference in depression. HA: As a result of 300mg/day of the ABC drug, there will be a significant difference in depression. H0: As a result of 300mg./day of the ABC drug, there will be a significant difference in depression. HA: As a result of 300mg/day of the ABC drug, there will be no significant difference in depression. Question 6 Match the null hypothesis (H 0 ) to the correct alternative hypothesis (H 1 ): __3__ H 0 : Lot A tensile strength is comparable to lot B tensile Strength __1__ H 0 : Lot A tensile strength exceeds or equals lot B tensile strength __2__ H 0 : Lot A tensile strength is not greater than lot B tensile strength 1 . H 1 : The mean of Lot A tensile strength is < the mean of Lot B tensile strength. 2 . H 1 : The mean of Lot A tensile strength is > the mean of Lot B tensile strength. 3 . H 1 : The mean of Lot A tensile strength ≠ the mean of Lot B tensile strength
Question 7 H0: Prototype design has at most 37 mpg vs. HA: Prototype design has greater than 37 mpg. If H0 is rejected, the action will be move the prototype design to production. What kind of test is required? Question options: A one-tailed test with lower reject region A one-tailed test with upper reject region A two-tailed test with lower and upper reject regions Question 8 Match up the following: __5__ The probability of a type 2 error __2__ Not reject H0 when H0 is false __6__ The power of a test = P(rejecting H0|H0 false) __4__ The risk we are willing to take of a type 1 error, or the type 1 error rate __1__ Reject H0 when H0 is true __3__ Reject H0 when H0 is false, or not reject H0 when H0 is not false 1 . Type 1 error 2 . Type 2 error 3 . Not an error 4 . α 5 . β 6 . 1 - β Question 9 For a test of the population proportion, what is the distribution of the test statistic? Question options: F T Z Question 10 H0: μ ≤ 16.92 vs. HA: μ > 16.92 What is the test statistic for sample of size 27, mean 13.90, and standard deviation 2.95? Enter the test statistic with 2 decimal places.
Answer: -5.32 Question 11 8 / 8 points A sample of size 20 yields a sample mean of 23.5 and a sample standard deviation of 4.3. Test H0: Mean ≥ 25 at α = 0.10. HA: Mean < 25. This is a one -tailed test with lower reject region bounded by a negative critical value. Question options: None of the other answers are correct. Pvalue 0.932. H0 not rejected. Conclude mean ≥ 25 plausible Pvalue 0.135. H0 not rejected. Conclude mean ≥ 25 plausible Pvalue 0.068. H0 rejected. Conclude mean < 25 Hide question 11 feedback Follow slides 29 and 30 to find the p value in the PPT for this week. Convert the x value to the t value. t = (23.5 - 25) / (4.3/sqrt(20)) = -1.56 (rounded) Next use this value to find the p value. Use =T.DIST(-1.56,19,1) = 0.0676 (rounded) Question 12 The results of sampling independent populations: sample 1 from population 1 mean 80 population variance 3
sample size 25 sample 2 from population 2 mean 81 population variance 2 sample size 50 Note that the population variances are known. Test H0: (population1 mean - population2 mean) ≤ 0 at α = 0.05. HA: (population1 mean - population2 mean) > 0. This is a one -tailed test with upper reject region and positive critical value. Question options: Pvalue 0.9938. H0 not rejected. Conclude difference of means ≤ 0 is plausible Pvalue 0.0124. H0 rejected. Conclude difference of means > 0. Pvalue 0.0062. H0 rejected. Conclude difference of means > 0. Hide question 12 feedback Since you know the population variance, The equation, from slide 67, to find the z value is =(80-81)/(sqrt(3/25 + 2/50)) = -2.5 (The problem gives the variance of each population. The variance is the square of the Standard Deviation, therefore, do not square the variance in the formula.) Next find p using =1-NORM.S.DIST(-2.5,1) = 0.994 (rounded) Since this is an upper tail test, use 1-NORM.S.DIST Question 13 The results of sampling independent populations: sample 1 from population 1 mean 1000 sample standard deviation 400 sample size 50 sample 2 from population 2
mean 1250 sample standard deviation 500 sample size 75 Test the H0: population1 mean = population2 mean at α = 0.01. HA: population1 mean ≠ population2 mean. This is a two-tailed test with both a negative lower critical value and a positive upper critical value. Separate variances is assumed. Question options: Test statistic of -3.09 < Critical value of - 1.29. H0 rejected. Conclude difference of means ≠ 0. Test statistic of -65.28 > Critical value of -1.29. H0 rejected. Conclude difference of means = 0 Test statistic of -3.09 < Critical value of - 2.62. H0 rejected. Conclude difference of means ≠ 0. Test statistic of 65.28 > Critical value of -2.36. H0 is not rejected. Conclude difference of means = 0. Hide question 13 feedback Use slide 61. T = (1000-1250)/sqrt(400^2/50+500^2/75) = -3.093 (rounded) This number is the test statistic. The degrees of freedom = 74+49 = 123 Now find the critical value. Use =T.INV(0.005, 123) = -2.616. (The probability is halved since this is a two tail problem.) The test statistic is less than the critical value. Therefore the test statistic is in the area of rejection. Do not accept the null hypothesis.