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STP 226 Chapter 10 practice Make sure to include all of the following elements: State the null and alternative hypotheses. Determine and compute the appropriate test statistics. Determine the rejection region or compute the p-value (as requested) Decide if the null hypothesis should be rejected, explain why and what it means for alternative hypothesis State your conclusion in words, answer question posed in the problem. 1. A "Get Smaller" weight loss program conducted a preliminary study to test the effectiveness of their program. 9 randomly selected clients had their weight recorded before and after 4 weeks of dieting. Assume that the difference between the before and after weights is normally distributed. Their respective weights in pounds are given in the table below: Client Before After 1 167 163 2 184 182 3 144 145 4 138 129 5 132 131 6 198 195 7 219 213 8 154 156 9 127 123 a) Perform a test at the = .05 level to determine if the program results in a mean loss of weight. b) Perform a test at the = .01 level to determine if the program results in a mean loss of weight. 2) Scientists studying the Australian Barking Mouse claim that the mean weight is different than that of the Chinese Barking Mouse. From prior studies, we can assume that the two species’ populations are normally distributed and have the same standard deviation. Two random samples (in grams) are given below Australian 34, 35, 41, 43, 45, 45, 48, 51 Chinese 29, 31, 32, 35, 44, 49 a) Perform an appropriate test to test the claim at the = .10 level. b) Find a 95% confidence interval for the difference between the weight of the Australian mouse and the Chinese Mouse. 3) Farmer Jones claims that his tomatoes mean weight is higher than that of farmer Smith’s. A random sample of 32 Jones tomatoes yielded a mean weight of 12.5 ounces and a standard deviation of 1.2 ounces, and a random sample of 35 Smith tomatoes yielded a mean weight of 11.9 ounces and a standard deviation of 1.6 ounces. a) Test this claim at the = .05 level. (use your calculator) b) Find a 99% confidence interval for the difference between the Jones and Smith weights. Solutions 1) H0: b a , Ha: b a d = 2.8889, sd = 3.4075, d.f. = 8, Crit Value t = 1.860, Test stat t = 2.543 p-value = .0173 (calculator T-Test) “At the .05 level, there is significant evidence to infer that people that after 4 weeks of the program, the clients’ mean weight decreases.” (for part b) Crit val t = 2.896 so do not reject H0. “At the .01 level, there is not sufficient evidence to that the mean weight has decreased.” 2a) Ho: A C , Ha: A C d.f. = 12, Crit values t = -1.782, 1.782, Test Stat t = 1.639 p-value = .1272 (calc) “At the .10 level, there is not sufficient evidence that the mean weights are different” b) (-2.004, 14.171) 3a) Ho: J S , Ha: J S p-value = .0429 “At the .05 level, there is sufficient evidence that Jones’ Tomatoes have a mean weight higher that that of Smith’s” b) (-.3131, 1.5131)