Problem # 5: (a) Erythromycin 1s a drug that has been proposed to possibly lower

Question

Problem # 5: (a) Erythromycin 1s a drug that has been proposed to possibly lower the risk of premature delivery. A related area of interest is its association with the incidence of side effects during pregnancy. Assume that 30% of all pregnant women complain of nausea between the 24th and 28th week of pregnancy. Furthermore, suppose that of 186 women who are taking erythromycin regularly during this period, 66 complain of nausea. Find the p-value for testing the hypothesis that incidence rate of nausea for the erythromycin group is greater than for a typical pregnant woman. (b) At the 2% significance level, what is the conclusion of the above hypothesis test? Problem #5(a): 0.9484 p-value (correct to 4 decimals) (A) We conclude that the incidence rate of nausea for the erythromycin group is greater than for a typical pregnant woman since the p-value is less than.01 (B) We cannot conclude that the incidence rate of nausea for the erythromycin group is greater than for a typical pregnant woman since the p-value is greater or equal to.01 (C) We cannot conclude that the incidence rate of nausea for the erythromycin group is greater than for a typical pregnant woman since the p-value is greater or equal to 0.02 (D) We cannot conclude that the incidence rate of nausea for the erythromycin group is greater than for a typical pregnant woman since the p-value is less than 0.02 (E) We cannot conclude that the incidence rate of nausea for the erythromycin group is greater than for a typical pregnant woman since the p-value is less than .01 (F) We conclude that the incidence rate of nausea for the erythromycin group is greater than for a typical pregnant woman since the p-value is greater than or equal to 0.02 (G) We conclude that the incidence rate of nausea for the erythromycin group is greater than for a typical pregnant woman since the p-value is less than 0.02 (H) We conclude that the incidence rate of nausea for the erythromycin group is greater than for a typical pregnant woman since the p-value is greater than or equal to .01 Problem #5(b): | F , conclusion Just Save Submit Problem #5 for Grading Problem #5 | Attempt #1 | Attempt #2 | Attempt #3 Your Answer: 5(a) 0.5438 5(a) 0.94845(a) Your Mark: 5(a) 0/2X 5(b) 0/2X 5(b) F 5(a) 0/2)x 5(b) 0/2X 5(a) 5(b)

Explanation / Answer

Solution:-

a)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P < 0.30
Alternative hypothesis: P > 0.30

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.02. The test method, shown in the next section, is a one-sample z-test.

Analyze sample data. Using sample data, we calculate the standard deviation (S.D) and compute the z-score test statistic (z).

S.D = sqrt[ P * ( 1 - P ) / n ]

S.D = 0.0336
z = (p - P) / S.D

z = 1.632

where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.

Since we have a one-tailed test, the P-value is the probability that the z-score is greater than 1.632.

Thus, the P-value = 0.051

Interpret results. Since the P-value (0.051) is greater than the significance level (0.02), we have to accept the null hypothesis.

b)

At the 2% significance level, we can conclude that we do not have sufficient evidence in the favor of the claim that incidence rate for nausea for this group is greater than for a typical pregnant women.

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