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Unformatted text preview: Hypothesis Testing Proportion and One Mean Solutions Pvalue Guidelines when using Standard Normal Table (i.e. the Ztable): Keep this in mind: The method for finding the pvalue is based on the alternative hypothesis. Minitab will provide the pvalue but if doing by hand using Table A1 observe the following: For Ha: p p o then the pvalue = 2P(Z z) That is, find 1 P(Z < z) and then multiply this pvalue by 2. For Ha: p > p o then the pvalue = P( Z z) For Ha: p < p o then the pvalue = P( Z z) 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 . H : 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 p = 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 pvalue? 0.019 d. Decide between the null hypothesis and the alternative hypothesis. Explain your decision. Decide on alternative hypothesis because pvalue 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 p = success rate for this sample? How does it compare to the success rate for the sample used in parts be?...
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This note was uploaded on 08/01/2011 for the course STAT 101 taught by Professor Thomas during the Spring '11 term at Pennsylvania State University, University Park.
 Spring '11
 Thomas

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