Question

Researchers studying osteoporosis (bone loss) suspected that women over the age of 50 in the United

States are diagnosed with the disease more often than women over 50 in Mexico. They took a random
sample of 200 women over the age of 50 from each country. Here are the results:
Diagnosed with osteoporosis? US. Mexico
Yes. 40. 20
No 160 180
Total 200. 200
The researchers want to use these results to test He: pus - PM = 0 versus H₂: Pus-PM > 0.
Assume that all conditions have been met.
What is the P-value associated with these sample results?
a. P-value is greater than or equal to
0.20
b. 0.05 is less than or equal to the P-
value < 0.10
c. 0.10 is less than or equal to the P-
value < 0.20
d. P-value < 0.01
e. 0.01 is less than or equal to the P-
value < 0.05

Answer 1

A P-value < 0.01

Step-by-step explanation:

Answer 2

The p-value associated with these sample results b. is less than 0.05, indicating that there is sufficient evidence to suggest that the percentage of women over 50 diagnosed with osteoporosis is higher in the United States than in Mexico.

Step-by-step explanation:

To find the p-value associated with the sample results, we first need to calculate the test statistic. The test statistic for a test of proportions is calculated using the formula:

Z = (p - p₀) / sqrt((p₀(1-p₀))/n)

where p is the sample proportion, p₀ is the hypothesized population proportion, and n is the sample size. In this case, p = 40/200 = 0.2, p₀ = 20/200 = 0.1, and n = 200. Plugging these values into the formula, we get:

Z = (0.2 - 0.1) / sqrt((0.1(1-0.1))/200) = 2

Next, we need to find the p-value corresponding to a test statistic of 2. Using a standard normal distribution table, we can find that the p-value is approximately 0.0228.

Since the p-value is less than 0.05, we reject the null hypothesis and conclude that there is sufficient evidence to suggest that the percentage of women over 50 diagnosed with osteoporosis is higher in the United States than in Mexico.