sol_hw_07

# sol_hw_07 - Solutions Hypothesis Testing 9.4 Iowa GPA H 0 =...

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Solutions - Hypothesis Testing 9.4 Iowa GPA : 0 H : μ= 2.80 a H : μ 2.80 In the above hypotheses, 0 H : is the notation for the null hypothesis, a H : is the notation for the alternative hypothesis, and μ is the parameter, the mean GPA of the population, about which we’re 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 0 H : p = 0.50 and a H : p > 0.50 b) p ˆ = 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. b)

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## This note was uploaded on 10/08/2010 for the course STAT 200 at Penn State.

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sol_hw_07 - Solutions Hypothesis Testing 9.4 Iowa GPA H 0 =...

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