Question

Regardless of age, about 20% of American adults participate in fitness activities at least twice a week. A random sample of 100 adults over 40 years old found 15 who exercised at least twice a week. Is this evidence of a decline in participation after age 40? Use α = .05. a) Write test hypothesis b) Compute test statistic. c) Apply any method to draw conclusion.

Answer 1

The null hypothesis is
`H_0: p = 0.2`

The alternate hypothesis is
`H_a: p < 0.2`

The p-value of the test is 0.1056 > 0.05, which means that there is not evidence of decline in participation after age 40 using α = 0.05.

Explanation:

20% of American adults participate in fitness activities at least twice a week. Test if there is evidence of a decline in participation after age 40.

At the null hypothesis, we test that the proportion is of 0.2, so:

`H_0: p = 0.2`

At the alternate hypothesis, we test if the proportion is less than 0.2, indicating a decline. So

`H_a: p < 0.2`

The test statistic is:

`z = (X - \mu)/((\sigma)/(√(n)))`

In which X is the sample mean,
`\mu` is the value tested at the null hypothesis,
`\sigma` is the standard deviation and n is the size of the sample.

20% is tested at the null hypothesis:

This means that
`\mu = 0.2, \sigma = √(0.2*0.8)`

A random sample of 100 adults over 40 years old found 15 who exercised at least twice a week.

This means that
`n = 100, X = (15)/(100) = 0.15`

Test statistic:

`z = (X - \mu)/((\sigma)/(√(n)))`

`z = (0.15 - 0.2)/((√(0.2*0.8))/(√(100)))`

`z = -1.25`

P-value of the test:

The p-value of the test is the probability of finding a sample proportion below 0.15, which is the p-value of z = -1.25.

Looking at the z-table, z = -1.25 has a pvalue of 0.1056.

The p-value of the test is 0.1056 > 0.05, which means that there is not evidence of decline in participation after age 40 using α = 0.05.