Answer to Question #203140 in Statistics and Probability for yoonji

Question

A certain health magazine stated that 14% of men said they used exercise to reduce stress. At a = 0.05, test the claim wherein, from a random sample of 100 men who were selected, 10 said that they used exercise to relieve stress.

Answer 1

$x=10 \\ n=100 \\ \hat{p}=\frac{x}{n} = \frac{10}{100}=0.1 \\ H_0: p=0.14 \\ H_1: p≠0.14$

Test statistic:

$Z = \frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}} \\ = \frac{0.1-0.14}{\sqrt{\frac{0.14(1-0.14)}{100}}} \\ = -1.15$

P-value for a two tailed z-test using standard normal distribution table

P-value=0.25

As P-value is greater than the level of significance, we do not have sufficient evidence at 5% to reject null hypothesis.

We have sufficient evidence to say that 14 % of men said they used exercise to reduce stress.