This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: Lesson 7 Homework Solutions 9.4 Iowa GPA : H : = 2.80 a H : 2.80 In the above hypotheses, H : is the notation for the null hypothesis, a H : is the notation for the alternative hypothesis, and is the parameter, the mean GPA of the population, about which were hypothesizing. 9.8 P-value: a) This P-value does not give strong evidence against the null hypothesis. b) This extreme P-value does give strong evidence against the null hypothesis. 9.17 Another test of therapeutic touch : a) p = proportion of trials guessed correctly H : p = 0.50 and a H : p > 0.50 b) p = 53/130 = 0.4077 se = ( 29 =- n p p / 1 ( 29 =- 130 / 50 . 1 50 . 0.0439 z = (0.4077 0.5)/0.0439 = - 2.10; the sample proportion is a bit more than 2 standard errors less than would be expected if the null hypothesis were true. c) The P-value is 0.9821 (rounds to 0.98). Because the P-value is so much greater than the significance level of 0.05, we do not reject the null hypothesis. The probability would be 0.98 of getting a test statistic at least as extreme as the value observed if the null hypothesis were true, and the population proportion were 0.50. d) np = (130)(0.5) = 65= n (1- p ); the sample size was large enough to make the inference in (c). We also would need to assume randomly selected practitioners and subjects for this to apply to all practitioners and subjects. 9.21 Garlic to repel ticks : a) The relevant variable is whether garlic or placebo is more effective, and the parameter is the population proportion, p = those for whom garlic is more effective than placebo....
View Full Document
- Spring '11