Download Stat 20, Fall 2005 - Homework #9 Solutions: Hypothesis Testing and Confidence Intervals and more Assignments Statistics in PDF only on Docsity! Stat 20, Fall 2005 Homework #9 solution 1 7.10 a. Weight changes are given in the table below: Subject 1 2 3 4 5 6 7 8 Weight change 6.0 3.9 6.4 3.9 4.8 6.7 3.6 6.0 Subject 9 10 11 12 13 14 15 16 Weight change 5.8 2.6 5.3 3.2 1.4 7.1 6.5 2.5 b. Mean weight change is x̄ = 4.73125 and standard deviation for the weight change is s = 1.745745. c. Computing the 95% confidence interval for the mean of the population, we find the standard error SEx̄ = s√ n = 1.745745√ 16 = 0.4364362 and margin of error for 95% c.i. is (sample size is 16) m = t∗SEx̄ = 2.131 · 0.4364362 = 0.9300456. Therefore 95% confidence interval for the mean of the population is (x̄ − m, x̄ + m) = (3.801204, 5.661296). We are 95% confident that the mean weight change is between 3.801204 kg and 5.661296 kg. d. Mean weight gain in pounds is x̄ = 10.40875 and standard deviation of the weight change in pounds is s = 3.840639. The 95% confidence interval for weight change in pounds is (x̄ − m, x̄ + m) = (8.36265, 12.45485). e. Null and alternative hypothesis are H0 : µ = 16 Ha : µ 6= 16 Stat 20, Fall 2005 Homework #9 solution 2 We fix significance level at α = 0.05. The t-statistic is t = x̄ − µ s/ √ n = 10.40875 − 16 3.840639/ √ 16 = −5.82325. Number of degrees of freedom here is 15, therefore P -value is P -value = 2P(T15 > |t|) = 3.354795 · 10−5. We see that P -value < α = 0.05, therefore we reject H0 at α = 0.05 significance level. f. Results we obtained with confidence interval are consistent with those obtained with hypothesis testing: confidence interval doesn’t cover point 16 lb, and we re- ject H0 at α = 0.05 significance level. Therefore we may conclude that hypothesis that 3500 extra calories translate into a weight gain of 1 pound doesn’t hold. 7.16 t-statistic is computed the same in case of one-sided test and two-sided test, it doesn’t change. The P -value for two-sided test is computed as P -value = 2P(T > |t|). Since x̄ is positive, then t-statistic value is also positive and therefore we can replace |t| with t in the equation above: P -value = 2P(T > t). At the same time P -value for one-sided test is computed as P -value = P(T > t). Therefore P -value for one-sided test is 0.04. 7.18 a. t∗ = 2.093. b. t∗ = 1.699. c. Since there is no row for df = 49, use row for df = 40: t∗ = 1.303. 7.20 a. df = 14. b. 2.145 < t < 2.264. c. 0.025 and 0.02. d. 0.02 < P -value < 0.025. e. See graph below.
