Microsoft Word - D XXXXXXXXXXand 3.1.docx Page 1 of 2 – Discussion 11 – Section 2.3 and 3.1 Discussion 11 – Sections 2.3 and 3.1 – 15 Points • You should answer the questions below in the discussion...

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Microsoft Word - D - 11 2.3 and 3.1.docx Page 1 of 2 – Discussion 11 – Section 2.3 and 3.1 Discussion 11 – Sections 2.3 and 3.1 – 15 Points • You should answer the questions below in the discussion area. • Make sure you label each answer with the appropriate number and letter so that students know what question you are addressing. • Leave a space between each answer. 1. In Exploration 1.1, we looked at a study to investigate whether Harley the dog could correctly choose between two cups when a researcher bowed toward one of the cups. In other words, the research question was about whether Harley could select the correct cup more than 50% of the time in the long run. a. Describe in context what a Type I error (false alarm) would be in this study. b. Describe in context what a Type II error (missed opportunity) would be in this study. 2. Suppose that we want to test H0: π = 0.74 vs. Ha: π ≠ 0.74. Let π denote some population proportion of interest and suppose a 99% confidence interval for π is calculated to be (0.6, 0.7). What can you say about the corresponding p-value? Explain your answer clearly so that it will be evident how you are thinking about this. 3. Suppose we are constructing a confidence interval using repeated tests of significance to develop an interval of plausible values for the population proportion, π. Using two-sided tests each time with the following null hypotheses, we obtain the p-values shown in the table on the next page. a. Using the results from the table given on the next page, give a 90% confidence interval for π . Briefly explain how you come up with the interval. b. Using the results from the table on the next page give a 95% confidence interval for π . Briefly explain how you come up with the interval. Page 2 of 2 – Discussion 11 – Section 2.3 and 3.1 Null p-value π = 0.45 0.014 π = 0.46 0.032 π = 0.47 0.062 π = 0.48 0.126 π = 0.49 0.371 π = 0.50 0.598 π = 0.51 0.733 π = 0.52 0.986 π = 0.53 0.787 π = 0.54 0.572 π = 0.55 0.373 π = 0.56 0.142 π = 0.57 0.077 π = 0.58 0.042 π = 0.59 0.021 π = 0.60 0.003