You are conducting a study to see if the proportion of women over 40 who regularly have mammograms is significantly less than 0.26 . You use a significance level of $\alpha=0.01$.
\[
\begin{array}{l}
H_{0}: p=0.26 \\
H_{1}: p< 0.26
\end{array}
\]
You obtain a sample of size $n=159$ in which there are 24 successes.
What is the test statistic for this sample? (Report answer accurate to three decimal places.) test statistic $=-3.135$ of
What is the $p$-value for this sample? (Report answer accurate to four decimal places.) p-value $=$

Step 1 ：State the null and alternative hypotheses. The null hypothesis is that the proportion of women over 40 who regularly have mammograms is 0.26, and the alternative hypothesis is that the proportion is less than 0.26. In mathematical terms, this is written as: \[H_{0}: p=0.26\] \[H_{1}: p<0.26\]

Step 2 ：Calculate the sample proportion, \(\hat{p}\), by dividing the number of successes by the sample size. In this case, there are 24 successes in a sample of 159, so \(\hat{p} = \frac{24}{159} = 0.1509433962264151\).

Step 3 ：Calculate the test statistic, Z, using the formula: \[Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}\] where \(\hat{p}\) is the sample proportion, \(p_0\) is the hypothesized population proportion, and n is the sample size. Substituting the given values, we get \[Z = \frac{0.1509433962264151 - 0.26}{\sqrt{\frac{0.26(1-0.26)}{159}}} = -3.1350754494877946\].

Step 4 ：Calculate the p-value, which is the probability of observing a test statistic as extreme as the one calculated, assuming the null hypothesis is true. In this case, we are looking for the probability of observing a test statistic less than -3.135 under the standard normal distribution. The p-value is 0.0008590495828120298.

Step 5 ：Round the test statistic and p-value to the required number of decimal places. The test statistic is rounded to three decimal places and the p-value is rounded to four decimal places. Therefore, the test statistic is \(-3.135\) and the p-value is \(0.0009\).

Step 6 ：Final Answer: The test statistic for this sample is \(\boxed{-3.135}\) and the p-value for this sample is \(\boxed{0.0009}\).