Name: Date: Hypothesis Testing Answer each question completely to recieve full credit 1. There is a new drug that is used to treat leukemia. The following data represents the remission time in weeks for a random sample of 21 patients using the drug. 10 11 10 7 20 34 32 19 32 23 6 25 22 17 13 6 35 9 16 6 6 Let X be a random variable representing the remission time in weeks for all patients using the new drug. Assume that the distribution of x is normal. A previously used drug treatment has a mean remission time of 12.5 weeks. Does the data indicate that the mean remission time using the new drug is different from 12.5 week at a level of significance of 0.01? State the Null Hypothesis Ho: There is no difference between the remission of the the new drug and previous drug. Variable 1 10 11 10 7 20 34 32 19 32 23 6 25 22 17 13 6 35 9 16 6 6 Alpha value (for confiden Variable #1 (Variable 1) Count Mean Mean LCL Mean UCL Variance Standard Deviation Mean Standard Error Coefficient of Variation Minimum Maximum Range State the Test Statistic: Two-tailed T-test Perform Calculations Statistical Conclusion Experimental Conclusion 6 35 Geometric Mean 29 Harmonic Mean Mode Median Median Error Percentile 25% (Q1) Percentile 75% (Q3) IQR MAD (Median Absolute D Coefficient of Dispersio State the Alternative Hypothesis State the Level of significance: 0.01 21 Mean Deviation 17.09524 Second Moment 12.54351 Third Moment 21.64697 Fourth Moment 99.99048 9.99952 Sum 2.18207 Sum Standard Error 0.58493 Total Sum Squares Adjusted Sum Squares Mean remission will still be 12. 5 H1: There is a difference in the mean remission time using the new drug and the mean remission time of the previous drug. 5.% sample size: 21 Mean: 17.09524 Median: 16 Variance: 99.99048 SD: 9.99048 16 0.59679 Skewness 9 Skewness Standard Erro 23 Kurtosis 14 Kurtosis Standard Error 10 Skewness (Fisher's) 0.52083 Kurtosis (Fisher's) 8.39456 95.22902 476.60037 17,664.49344 359 45.82357 8,137 1,999.80952 14.31179 11.91937 6 0.51286 0.47673 1.94788 0.83101 0.55319 -1.00273 Hypothesis Testing Answer each question completely to recieve full credit. 2. We wish to test the claim that the mean body mass index (BMI) of men Men is equal to the mean BMI of women. Use the data to the right to test this claim. 20 37 46 23 20 23 21 15 20 28 27 20 30 22 27 38 29 20 16 27 42 37 39 39 32 16 21 26 17 39 State the Null Hypothesis: Ho: BMI of men is equal to the BMI of women State the Alternatice Hypothesis: H1:BMI of men is not equal to the BMI of women State the Level of significance: Level of Significance is 0.05 State the Test Statistic: T-test - Two Sample Perform Calculations Mean for Men= 27.2333 SD= 8.78354 Mean for Women= 30.4333333 SD=10.59825 Calculations: (See Box - Comparing Means) P-level: 0.20799 > Significance level: 0.05 P>0.05 Statistical Conclusion P>0.05 Failed to reject the Null hypothesis. Experimental Conclusion We failed to reject the null hypothesis that the mean BMI of men is equal to the mean BMI of women. Women Mean Standard Deviation 29 28 20 28 42 45 19 45 16 32 38 45 41 34 28 21 42 21 30 28 30 43 40 16 44 15 16 20 41 16 27.2333333333 8.78354 30.4333333333 10.59825 Comparing Means [ t-test assuming equal variances (homoscedastic) ] Descriptive Statistics VAR Men Women Sample size Mean 30 30 Variance 27.23333 30.43333 77.15057 112.32299 Summary Degrees Of Freedom Test Statistics 58 Hypothesized Mean Differenc 1.27332 Pooled Variance 0 94.73678 Two-tailed distribution p-level 0.20799 t Critical Value (5%) 2.00172 One-tailed distribution p-level 0.10399 t Critical Value (5%) 1.67155 G-criterion Test Statistics Critical Value (5%) 0.10492 p-level 0.143 0.08283 Pagurova criterion Test Statistics Ratio of variances para 1.27332 p-level 0.40718 Critical Value (5%) 0.79183 0.06299