Solutions_Activity_10
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Solutions_Activity_10

Course Number: STAT 200, 2009

College/University: Penn State

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ACTIVITY 10 - Solutions P-value Guidelines: Keep this in mind: The method for finding the p-value is based on the alternative hypothesis. Minitab will provide the p-value but if doing by hand using Table A1 observe the following: For Ha: p po then p-value = 2P(Z |z|) For Ha: p > po then p-value = P( Z z) For Ha: p < po then p-value = P( Z z) Activity 10.1 The present success rate in the treatment of a...

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10 ACTIVITY - Solutions P-value Guidelines: Keep this in mind: The method for finding the p-value is based on the alternative hypothesis. Minitab will provide the p-value but if doing by hand using Table A1 observe the following: For Ha: p po then p-value = 2P(Z |z|) For Ha: p > po then p-value = P( Z z) For Ha: p < po then p-value = P( Z z) Activity 10.1 The present success rate in the treatment of a particular psychiatric disorder is 0.65 (65%). A research group creates a new treatment for this disorder. Their null hypothesis is that the success rate for the new treatment is 0.65 (no different from the standard). The alternative hypothesis is that the success rate is better than 0.65 for the new treatment. a. Let p = true success rate of the new treatment. Using mathematical notation, write null and alternative hypotheses about p. H0: p = .65 versus Ha: p > .65. b. A clinical trial is done in which 144 of 200 patients who use the new treatment are successfully treated. What is the value of = success rate for the sample? How does it compare to 0.65 (the old standard)? 144/200 = .72. Sample value is greater than .65. c. In Minitab use Stat>Basic Stats>1 proportion, click Summarized Data, enter 200 for number of trials and 144 for Number of events. Click on Options, AND enter .65 where it says Test proportion AND select the alternative hypothesis as greater than AND also click on Use test and interval based on normal distribution. What value is given for the test statistic Z in the output? 2.08 What is the p-value? 0.019 d. Decide between the null hypothesis and the alternative hypothesis. Explain your decision. Decide on alternative hypothesis because p-value is less than .05. e. Write a conclusion about how the new treatment compares to previous treatment(s). The new treatment appears to have a better success rate. The sample result was a statistically significant departure from p = .65. 1 f. Suppose the data had been that 50 patients used the new treatment, with 36 successes. What is the value of = success rate for this sample? How does it compare to the success rate for the sample used in parts be? Sample result is still .72 (same as before). g. With the data given in part f, use Minitab to do a hypothesis test of whether the true success rate for the new treatment is not or is greater than .65. That is, repeat part c but change the number of trials and events to 50 and 36 respectively. What value is given for the test statistic Z in the output? 1.04 Choose the null because the p-value is not less than .05. What is the p-value? 0.150 h. Refer to the previous two parts. Decide between the null hypothesis and the alternative hypothesis. Explain your decision. Choose the null because the p-value is not less than .05. We are not able to say that the new treatment is better than the previous treatment(s). i. For the trial with only 50 patients (and 36 success), write a conclusion about how the new treatment compares to previous treatment(s). We are not able to say that the new treatment is better than the treatment(s). j. previous Briefly explain what this activity illustrates about how sample size affects the statistical significance of an observed result. As a starting points, note that the observed success rate was .72 for both samples, and we wish to determine if this is significant evidence that the true proportion is greater than .65. A specific amount of difference between a sample result and a null hypothesis becomes more significant as the sample size increases. Activity 10.2 In a marketing survey for an automobile manufacturer, 90 randomly selected adults are asked which car color they would choose, if a particular car were available in either blue or red body colors. Of the 90 respondents, 53 said blue. a. Let p = population proportion that would say blue. The manufacturer wants to learn if a majority of buyers would pick blue. Keeping in mind that a majority is p>0.5, write a null and alternative hypothesis about p in this situation. (Hint: What somebody wants to prove is usually the alternative.) H0: p = .5 (or less) Hs: p> .5 b. What is the value of =sample proportion that picked blue? 53/90 = .59 2 c. Test the hypotheses stated in part a above. By hand, calculate the test statistic by using (Notice that this statistic is sensitive to the difference between the sample result and the null hypothesis value): = = 1.71 (round your final value to two decimal places) d. Use table A1 (either in your book or in the folder for this week) to find the p-value associated with this test statistic. Use the p-value guidelines found at the beginning of this activity. From Table A1, P(Z > 1.71) = 1 0.9564 = 0.0436 e. Use Minitab as you did for Activity 10.1, but youll have to change the numbers of trials and events AND use the Options button to change the null hypothesis value of p. What value is given for Z in the output? 1.69 What is the p-value? 0.046 i. Do the Z test statistic you found by hand in part c and the p-value from part d approximately equal to the Z statistic found in part e when using Minitab? Yes ii. Decide whether the result is significant based on the p-value from Minitab and report a conclusion in the context of this situation. With a p-value of 0.046 which is less than 0.05 we would reject the null hypothesis and conclude that our result is significant. The manufacturer would be correct in saying that a majority of buyers would pick blue M&Ms. iii What would the p-value have been if manufacturer wanted to test if a equal proportion of red and blue cars would be purchased? That is, test Ho: p = 0.5 versus Ha: p 0.5 For a two-sided test we would double the p-value found from a one-sided test resulting in a p-value of 0.092 which would be greater than 0.05 meaning we would not reject Ho for the two sided test. This helps (hopefully!) to illustrate the importance of selecting the correct alternative hypothesis, Ha. Depending on the selected Ha you can reach different decisions and conclusions. (NOTE: conversely, for a one-sided test we would use half of the p-value from a two-sided test.) 3

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