Question

A study claims that all adults spend an average of 14 hours or more on chores during a weekend. A researcher wanted to check if this claim is true. A random sample of 200 adults taken by this researcher showed that these adults spend an average of 14.65 hours on chores during a weekend. The population standard deviation is known to be 3.0 hours.
a. Find the p-value for the hypothesis test with the alternative hypothesis that all adults spend more than 14 hours on chores during a weekend. Will you reject the null hypothesis at α = .05? b. Test the hypothesis of part a using the critical-value approach at α = .05.

Answer 1

Answer:

a) The null and alternative hypotheses:

H0 : u ≤ 14

H1 : u > 14

This is a right tailed test.

Let's find test statistics, z

$Z = \frac{x' - u}{\sigma / \sqrt{n}}$

$Z = \frac{14.65 - 14}{3/ \sqrt{200}} = 3.0641$

P-value:

Using the standard normal table, NORMSDIST(3.0641) = 0.99889

P(Z>3.0641) = 1 - 0.99889

P(Z>3.0641) = 0.00111

Pvalue = 0.00111

At a significance level of 0.05.

Since pvalue, 0.00111 is less than significance level, 0.05, we reject null hypothesis, H0.

b) Test the hypothesis of part a using the critical-value approach at α = .05

At α = 0.05, (1 - 0.05) = 0.95

Using the standard normal table, the critical value at 0.95 is 1.65.

Reject null hypothesis H0, since Zstat 3.0641 is greater than Zcritical, 1.65.