Question

2.1.38 A zoologist at a large metropolitan zoo is concerned about a potential new disease present among the 243 sharks living in the large aquarium at the zoo. The zoologist takes a random sample of 15 sharks from the aquarium, temporarily removes the sharks from the tank and tests them for the disease. He finds that 3 of the sharks have the disease. The zoologist wishes to test whether there is evidence that less than one-fourth of the sharks in the aquarium are diseased. Evaluate the strength of evidence for this hypothesis. Find the p-value for the hypothesis using a simulation-based approach Based on the p-value evaluate the strength of evidence and state a conclusion about diseased sharks. In which population, if any, are you comfortable drawing your conclusion? Explain why a theory-based approach is or is not reasonable for these data. If a theory-based approach is reasonable, find the p- value. O Atheory-based approach is reasonable since the p-value is the same as the p-values found using a simulation-based approach. O Atheory-based approach is not reasonable since the p-value does not match the p-value found using a simulation- based approach O A theory-based approach is not reasonable since there were only 3 sharks with the disease. We need at least 10. O Atheory-based approach is reasonable since the sample size is less than 5% of the population.

Answer 1

The following information is provided: The sample size is N = 15, the number of favorable cases is X = 3, and the sample proportion is p = * = * = 0.2, and the significance level is a = 0.05 (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho:p= 0.25 Ha: p < 0.25 This corresponds to a left-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 left-tailed test is zc -1.64. = = The rejection region for this left-tailed test is R = {z :2 < -1.64}

(3) Test Statistics The Z-statistic is computed as follows: z p – Po vpo(1 – po)/n 0.2 - 0.25 ✓0.25(1 – 0.25)/15 -0.447 (4) Decision about the null hypothesis Since it is observed that z = -0.447 > Zc = -1.64, it is then concluded that the null hypothesis is not rejected. = 0.3274 > 0.05, it is Using the P-value approach: The p-value is p = 0.3274, and since p concluded that the null hypothesis is not rejected.

(5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population proportion p is less than po, at the a = 0.05 significance level. Confidence Interval The 95% confidence interval for p is: -0.002 < p < 0.402. Graphically. Z-Test Results: z-stats = -0.447, p-value =0.3274 0.40 0.35 0.30 0.25 0.20 0.15 0.10 0.05 0.00 -4.0 -3.5 -3.0-2.5 -2.0-1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

Theory base approach is not reasonable since the sample size is less than 5% of the population.

Also there were only three sharks with the disease. We need at least 10.

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