In an experiment to learn if substance M can help restore memory, the brains of
ID: 3246198 • Letter: I
Question
In an experiment to learn if substance M can help restore memory, the brains of 22 rats were treated to damage their memories. The rats were trained to run a maze. After a day, 11 rats were given M, and seven of them succeeded in the maze: only three of the 11 control rats were successful. The z test for "no difference" against "a higher proportion of the M group succeeds" has what values for the z-statistic and P-value? (Round your z- value to two decimal places and your P-value to three decimal places.) Z = P = You may need to use the appropriate Appendix Table to answer this question.Explanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: P1 = P2
Alternative hypothesis: P1 P2
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the proportion from population 1 is too big or if it is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a two-proportion z-test.
Analyze sample data. Using sample data, we calculate the pooled sample proportion (p) and the standard error (SE). Using those measures, we compute the z-score test statistic (z).
p = (p1 * n1 + p2 * n2) / (n1 + n2)
p = 0.455
SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }
SE = 0.2124
z = (p1 - p2) / SE
z = 1.71
where p1 is the sample proportion in sample 1, where p2 is the sample proportion in sample 2, n1 is the size of sample 1, and n2 is the size of sample 2.
Since we have a two-tailed test, the P-value is the probability that the z-score is less than - 1.71 or greater than 1.71.
Thus, the P-value = 0.0872
Interpret results. Since the P-value (0.0872) is greater than the significance level (0.05), we have to accept the null hypothesis.
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