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

Using a one-sample Z-test, the test statistic was calculated as 3.0833, which corresponds to a p-value much less than 0.01, leading to the rejection of the null hypothesis at α = 0.05. The critical value approach also supports rejecting the null hypothesis as the test statistic exceeds the Z-critical value of 1.645.

Step-by-step explanation:

To address the student's question regarding hypothesis testing for the average hours adults spend on chores during a weekend, we must first define the null and alternative hypotheses and then calculate the test statistic. Afterward, we find the p-value and make a decision based on the significance level, α = 0.05. The null hypothesis (H0) is that adults spend an average of 14 hours on chores (μ = 14), while the alternative hypothesis (Ha) is that they spend more than 14 hours on chores (μ > 14).

Given the information:

• Sample mean (μ) = 14.65 hours
• Population standard deviation (σ) = 3.0 hours
• Sample size (n) = 200

We use a one-sample Z-test to calculate the test statistic:

Z = (X - μ) / (σ / √n) = (14.65 - 14) / (3 / √200) = 3.0833

Using the Z-table, find the p-value corresponding to the Z-score of 3.0833. The p-value is very small (much less than 0.01), indicating strong evidence against the null hypothesis. Therefore, we reject the null hypothesis at α = 0.05 since the p-value is less than 0.05.

For part b), the critical value approach, we find the critical Z-value for a one-tailed test at α = 0.05, which is approximately 1.645. Since our test statistic of 3.0833 is greater than the critical value of 1.645, we again reject the null hypothesis.

In conclusion, there is sufficient evidence to support the claim that adults spend more than 14 hours on chores during a weekend.

Answer 2

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 = (x' - u)/(\sigma / √(n))`

`Z = (14.65 - 14)/(3/ √(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.