A survey by Men’s Health magazine stated that 14% of men said they used exercise to reduce stress. Use a = 0.10. a random sample of 100 men was selected and 10 said that they used exercise to relieve stress. Use the P –value method to test the claim. Could the results be generalized to all adult Americans?
To test the claim using the P-value method, we need to set up the null and alternate hypotheses. Let's define the null hypothesis (H0) as "the proportion of all adult American men who use exercise to reduce stress is equal to 14%," and the alternate hypothesis (Ha) as "the proportion of all adult American men who use exercise to reduce stress is not equal to 14%."
The next step is to calculate the test statistic. In this case, we'll use the z-test statistic for proportions. The formula for the z-test statistic is given by:
z = (p̂ - p0) / √(p0 * (1 - p0) / n)
Where p̂ is the sample proportion, p0 is the hypothesized proportion, and n is the sample size.
Given that the sample size (n) is 100 and the proportion of men who said they used exercise to relieve stress (p̂) is 10/100 = 0.10, we can calculate the z-test statistic:
z = (0.10 - 0.14) / √(0.14 * (1 - 0.14) / 100)
= -0.04 / √(0.14 * 0.86 / 100)
= -0.04 / √0.01204
= -0.04 / 0.1097
≈ -0.364
Next, we need to find the P-value associated with this test statistic. The P-value is the probability of observing a test statistic as extreme as the one calculated, assuming the null hypothesis is true.
Since this is a two-tailed test (we are testing if the proportion is not equal to 14%), we need to calculate the probability of observing a z-value less than -0.364 and greater than 0.364.
Using a standard normal distribution table or a calculator, we find that the probability of observing a z-value less than -0.364 is approximately 0.3594, and the probability of observing a z-value greater than 0.364 is also approximately 0.3594.
To find the total P-value, we sum both tail probabilities: 0.3594 + 0.3594 = 0.7188.
This P-value of 0.7188 is larger than the significance level (α) of 0.10. Therefore, we fail to reject the null hypothesis.
In conclusion, based on the given sample data and using the P-value method, we do not have sufficient evidence to conclude that the proportion of all adult American men who use exercise to reduce stress is different from 14%. However, we cannot definitively generalize these results to all adult Americans as the sample may not be representative of the entire population. Further research and exploration may be required to make broader conclusions.