Question 2, 8.2.18-T
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A genetic experiment involving peas yielded one sample of offspring consisting of 430 green peas and 135 yellow peas. Use a 0.05 significance level to test the claim that under the same circumstances, $27 \%$ of offspring peas will be yellow. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. Use the P-value method and the normal distribution as an approximation to the binomial distribution.
\[
H_{1}: p> 0.27
\]
$\mathrm{H}_{0}: \mathrm{p} \neq 0.27$
\[
H_{1}: p< 0.27
\]
\[
H_{1}: p< 0.27
\]
\[
\begin{array}{l}
H_{0}: p \neq 0.27 \\
H_{1}: p=0.27
\end{array}
\]
What is the test statistic?
\[
z=
\]
(Round to two decimal places as needed.)
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Answer
Final Answer: The test statistic is \(\boxed{-1.66}\).
Step 1 ：The null hypothesis is \(H_{0}: p = 0.27\) and the alternative hypothesis is \(H_{1}: p \neq 0.27\).
Step 2 ：The sample proportion \(\hat{p}\) is calculated as the number of yellow peas divided by the total number of peas, which is \(\frac{135}{565} = 0.239\).
Step 3 ：The hypothesized population proportion \(p_{0}\) is given as 0.27.
Step 4 ：The sample size \(n\) is the total number of peas, which is 565.
Step 5 ：The test statistic in a hypothesis test for a proportion is a z-score, which is calculated using the formula \(z = \frac{\hat{p} - p_{0}}{\sqrt{\frac{p_{0}(1-p_{0})}{n}}}\).
Step 6 ：Substituting the values into the formula, we get \(z = \frac{0.239 - 0.27}{\sqrt{\frac{0.27(1-0.27)}{565}}} = -1.66\).
Step 7 ：Final Answer: The test statistic is \(\boxed{-1.66}\).
