Erythromycin is a drug that has been proposed to possibly lower the risk of prem

Question

Erythromycin is 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 in a sample of women who are taking erythromycin regularly during this period, 63 complain of nausea.

Suppose that we use this sample to test the hypothesis that incidence rate of nausea for the erythromycin group is greater than for a typical pregnant woman, and find the p-value to be 0.019. What is the meaning of this p-value? (A) The p-value of 0.019 is the probability of making a wrong decision. (B) If Erythromycin does not increase the incidence of nausea
it is very unlikely (probability 0.019) to obtain exactly 63
complaints of nausea in the given sample. (C) The p-value of 0.019 is the probability of making a Type II error. (D) If Erythromycin does increase the incidence of nausea
it is very unlikely (probability 0.019) to obtain 63 or more
complaints of nausea in the given sample. (E) The p-value of 0.019 is the probability of making a Type I error. (F) If Erythromycin does increase the incidence of nausea
it is very unlikely (probability 0.019) to obtain exactly 63
complaints of nausea in the given sample. (G) If Erythromycin does not increase the incidence of nausea
it is very unlikely (probability 0.019) to obtain 63 or more
complaints of nausea in the given sample. Erythromycin is 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 in a sample of women who are taking erythromycin regularly during this period, 63 complain of nausea.

Suppose that we use this sample to test the hypothesis that incidence rate of nausea for the erythromycin group is greater than for a typical pregnant woman, and find the p-value to be 0.019. What is the meaning of this p-value? Erythromycin is 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 in a sample of women who are taking erythromycin regularly during this period, 63 complain of nausea.

Suppose that we use this sample to test the hypothesis that incidence rate of nausea for the erythromycin group is greater than for a typical pregnant woman, and find the p-value to be 0.019. What is the meaning of this p-value? (A) The p-value of 0.019 is the probability of making a wrong decision. (B) If Erythromycin does not increase the incidence of nausea
it is very unlikely (probability 0.019) to obtain exactly 63
complaints of nausea in the given sample. (C) The p-value of 0.019 is the probability of making a Type II error. (D) If Erythromycin does increase the incidence of nausea
it is very unlikely (probability 0.019) to obtain 63 or more
complaints of nausea in the given sample. (E) The p-value of 0.019 is the probability of making a Type I error. (F) If Erythromycin does increase the incidence of nausea
it is very unlikely (probability 0.019) to obtain exactly 63
complaints of nausea in the given sample. (G) If Erythromycin does not increase the incidence of nausea
it is very unlikely (probability 0.019) to obtain 63 or more
complaints of nausea in the given sample. (A) The p-value of 0.019 is the probability of making a wrong decision. (B) If Erythromycin does not increase the incidence of nausea
it is very unlikely (probability 0.019) to obtain exactly 63
complaints of nausea in the given sample. (C) The p-value of 0.019 is the probability of making a Type II error. (D) If Erythromycin does increase the incidence of nausea
it is very unlikely (probability 0.019) to obtain 63 or more
complaints of nausea in the given sample. (E) The p-value of 0.019 is the probability of making a Type I error. (F) If Erythromycin does increase the incidence of nausea
it is very unlikely (probability 0.019) to obtain exactly 63
complaints of nausea in the given sample. (G) If Erythromycin does not increase the incidence of nausea
it is very unlikely (probability 0.019) to obtain 63 or more
complaints of nausea in the given sample. Erythromycin is 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 in a sample of women who are taking erythromycin regularly during this period, 63 complain of nausea.

Suppose that we use this sample to test the hypothesis that incidence rate of nausea for the erythromycin group is greater than for a typical pregnant woman, and find the p-value to be 0.019. What is the meaning of this p-value? (A) The p-value of 0.019 is the probability of making a wrong decision. (B) If Erythromycin does not increase the incidence of nausea
it is very unlikely (probability 0.019) to obtain exactly 63
complaints of nausea in the given sample. (C) The p-value of 0.019 is the probability of making a Type II error. (D) If Erythromycin does increase the incidence of nausea
it is very unlikely (probability 0.019) to obtain 63 or more
complaints of nausea in the given sample. (E) The p-value of 0.019 is the probability of making a Type I error. (F) If Erythromycin does increase the incidence of nausea
it is very unlikely (probability 0.019) to obtain exactly 63
complaints of nausea in the given sample. (G) If Erythromycin does not increase the incidence of nausea
it is very unlikely (probability 0.019) to obtain 63 or more
complaints of nausea in the given sample.

Explanation / Answer

here for p value is probability of having test statistic as extreme or more given null is true

option G is correct

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