Question

Name: Section Number: To be graded, all assignments must be completed and submitted on the original book page Introduction It is important that you be able to compute correctly. However, computation prowess is no substitute for a deeper understanding of what you are doing and why. This is especially true in the field of statistical science. At the undergraduate level, computations are pretty easy. At all levels, however, the underlying concepts are challenging. The following activity will demonstrate and elucidate this divide. EXHIBIT 1 Eureka or Not? Able to Reject HP S 0.32? 1 No No No No No Suppose that twenty identical experiments are taking place simultaneously around the world. The researchers are all studying the same drug, which they hope will improve the survival rate of the black- winged peckerwood finch after it has been infected with a particular type of tree mold. The survival rate left untreated is unfortunately only 32%. None of the researchers know about the others' work. The table to the right shows the results from the 20 different studies. In all cases, the significance level was a = 0.05 and the hypothesis being tested was: 00 TABLE 5.3 Finch Survival Rates Observed Number of Site Survival Rate Finches Studied with Drug 0.35 n=100 2 0.34 n100 3 0.31 n100 4 0.33 n100 5 0.33 n100 6 0.35 n=100 7 0.35 n 100 0.33 n=100 9 0.30 n100 10 0.34 n100 11 0.34 n=100 12 0.30 n=100 13 0.34 n = 100 14 0.31 15 0.31 n = 100 16 0.31 ne 100 17 0.45 n=100 18 n100 19 0.35 n = 100 20 0.33 n=100 No No No No No No No No No No n=100 0.30 No Yes No No No He: p=0.32 H:p> 0.32 Theme 5: Hypothesis Testing &

Answer 1

1)

Ho:p= 0.32 Ha: p > 0.32 This corresponds to a right-tailed test, for which a z-test for one population proportion needs to be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the critical value for a right-tailed test is ze = 1.64. The rejection region for this right-tailed test is R = {z: z > 1.64} (3) Test Statistics The Z-statistic is computed as follows: 2= p - po Po(1 – po)/n 0.45 – 0.32 0.32(1 – 0.32)/100 2.787 (4) Decision about the null hypothesis Since it is observed that z = 2.787 > 2c = 1.64, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p=0.0027, and since p=0.0027 < 0.05, it is concluded that the null hypothesis is rejected. (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population proportion p is greater than po, at the a = 0.05 significance level.

p-value = P(Z > TS)   {for right-tailed test)

= P(Z > 2.787)

= 0.0027

since p-value < alpha, we reject the null hypothesis

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