Question

3) An article in the Archives of Internal Medicine reported that in a sample of 244 men, 73 had elevated total cholesterol levels (more than 200 miligrams per deciliter). In a sample of 232 women, 44 had elevated total cholesterol levels. We will assume that we have a sample that meets the criteria for a normal approximation.

a) Write the hypotheses for the test.

b) Calculate the difference in sample proportions between groups.

	Response	No Response
Treatment A	20	105
Treatment B	5	30

	Response	No Response
Treatment A	25	10
Treatment B	90	40

c) Using 2PropZTest, calculate a P-value.

d) Fill in the blanks to explain what the P-value means:
If there is __________________________ between the proportions of high cholesterol, then there is a ___ % chance that there is a _______ or ______________ percentage point difference between the high cholesterol rates.

e) What is your decision about the null hypothesis? Why?

f) Statistically, is there a significant difference between high cholesterol rates of men and women? Support your answer.

g) Calculate a 95% confidence interval for the difference in high cholesterol rates. Does the confidence interval strengthen orweaken your conclusions in parts e) and f)? Why?

Answer 1

1. Null hypothesis: The proportion of people with elevated cholesterol levels is the same between men and women

Alternate hypothesis: The proportion of people with elevated cholesterol levels differs between men and women

We use H0 to represent the null and HA to represent the alternative (sometimes people use H1). Note that the following are for testing proportions (you would substitute for p to test means.

2. Calculate the difference in sample proportions between groups.

Level of significance = 0.05

The z score test for two population proportions is used when you want to know whether two populations or groups (males and females) differ significantly on some single (categorical) characteristic - like elevated cholesterol levels

Test statistic: z test for 2 proportion

where,

pm = proportion of men with an elevated cholesterol level = 73/244 = 0.299

pw = proportion of women with an elevated cholesterol level = 44/232 = 0.1896

p = overall sample proportion; The numerator will be the total number of “positive” results for the two samples and the denominator is the total number of people in the two samples= (73+44)/(244+232) = 0.2457

nm = total number of samples of men = 244

nw = total number of samples of women = 232

c) Using 2PropZTest, calculate a P-value.

p-value = 0.0056

d) Fill in the blanks to explain what the P-value means:
If there is a difference between the proportions of high cholesterol, then there is a 99.44% chance that there is an observed or more percentage point difference between the high cholesterol rates.

e) What is your decision about the null hypothesis? Why?

Reject the null hypothesis. We have enough evidence to support the claim that that the proportion of people with elevated cholesterol levels differs between men and women with 95% confidence level because p-value of the statistic(0.0056) is smaller than the level of significance (0.05).

f) Statistically, is there a significant difference between high cholesterol rates of men and women? Support your answer.

yes, there is a significant difference between high cholesterol rates of men and women because the p-value of the statistic(0.0056) is smaller than the level of significance (0.05).

g) Calculate a 95% confidence interval for the difference in high cholesterol rates. Does the confidence interval strengthen or weaken your conclusions in parts e) and f)? Why?

The confidence interval of proportion is given by following formula:

For 95% confidence interval, z critical = 1.96

The difference in proportion will lie between

we can be 95% confident that between 3.26% to 18.54% more men will have elevated cholesterol than women. yes, it strengthens the claim made in other parts.