Problem

Mrs. Obama implemented a program to get "children playing outside" and predicts it increased the proportion of time children spent outside on the weekends to more than $15 \%$. To test this prediction, she surveyed 300 random family participants and found that children of 205 families spent more than $15 \%$ of their weekend time outside.
The following is the setup for this hypothesis test:
\[
\begin{array}{l}
H_{0}: p=0.15 \\
H_{a}: p> 0.15
\end{array}
\]
The $p$-value for this hypothesis test is 0.002 .
At the $2.5 \%$ significance level, should she reject or fail to reject the null hypothesis?

Answer

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Answer

Final Answer: Since the p-value is less than the significance level, we should \(\boxed{\text{Reject the null hypothesis}}\).

Steps

Step 1 :Mrs. Obama implemented a program to get 'children playing outside' and predicts it increased the proportion of time children spent outside on the weekends to more than 15%. To test this prediction, she surveyed 300 random family participants and found that children of 205 families spent more than 15% of their weekend time outside.

Step 2 :The following is the setup for this hypothesis test: \[\begin{array}{l} H_{0}: p=0.15 \\ H_{a}: p>0.15 \end{array}\]

Step 3 :The p-value for this hypothesis test is 0.002.

Step 4 :At the 2.5% significance level, we need to decide whether to reject or fail to reject the null hypothesis.

Step 5 :The p-value is the probability of obtaining the observed data (or data more extreme) if the null hypothesis is true. If the p-value is less than the significance level, we reject the null hypothesis.

Step 6 :In this case, the p-value is 0.002, which is less than the significance level of 0.025.

Step 7 :Final Answer: Since the p-value is less than the significance level, we should \(\boxed{\text{Reject the null hypothesis}}\).

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