Workers at a certain soda drink factory collected data on the volumes (in ounces) of a simple random sample of 25 cans of the soda drink. Those volumes have a mean of $12.19 \mathrm{oz}$ and a standard deviation of $0.11 \mathrm{oz}$, and they appear to be from a normally distributed population. If the workers want the filling process to work so that almost all cans have volumes between $11.97 \mathrm{oz}$ and $12.61 \mathrm{oz}$, the range rule of thumb can be used to estimate that the standard deviation should be less than $0.16 \mathrm{oz}$. Use the sample data to test the claim that the population of volumes has a standard deviation less than $0.16 \mathrm{oz}$. Use a 0.025 significance level. Complete parts (a) through (d) below.
a. Identify the null and alternative hypotheses. Choose the correct answer below.
A.
\[
\begin{array}{l}
H_{0}: \sigma=0.16 \mathrm{oz} \\
H_{1}: \sigma \neq 0.16 \mathrm{oz}
\end{array}
\]
C.
\[
\begin{array}{l}
H_{0}: \sigma \geq 0.16 \mathrm{oz} \\
H_{1}: \sigma< 0.16 \mathrm{oz}
\end{array}
\]
B.
\[
\begin{array}{l}
H_{0}: \sigma=0.16 \mathrm{oz} \\
H_{1}: \sigma< 0.16 \mathrm{oz}
\end{array}
\]
D.
\[
\begin{array}{l}
\mathrm{H}_{0}: \sigma> 0.16 \mathrm{oz} \\
\mathrm{H}_{1}: \sigma=0.16 \mathrm{oz}
\end{array}
\]
b. Compute the test statistic.
\[
\chi^{2}=11.344
\]
(Round to three decimal places as needed.)
c. Find the P-value.
\[
\text { P-value }=\square
\]
(Round to four decimal places as needed.)
Answer
The final answer is: The correct null and alternative hypotheses are \(H_{0}: \sigma=0.16 \mathrm{oz}\) and \(H_{1}: \sigma<0.16 \mathrm{oz}\). The test statistic is \(\boxed{11.344}\). The P-value is \(\boxed{0.9863}\).
Step 1 ：Identify the null and alternative hypotheses. The correct null and alternative hypotheses are: \(H_{0}: \sigma=0.16 \mathrm{oz}\) and \(H_{1}: \sigma<0.16 \mathrm{oz}\).
Step 2 ：Compute the test statistic. The test statistic is calculated using the formula for the chi-square statistic with the given sample size of 25, sample standard deviation of 0.11 oz, and population standard deviation under the null hypothesis of 0.16 oz. The calculated test statistic is \(\chi^{2}=11.344\).
Step 3 ：Find the P-value. The P-value is calculated using the chi-square distribution with 24 degrees of freedom (since the degrees of freedom is n-1). The P-value is the area to the right of the test statistic in the chi-square distribution, since we are testing the alternative hypothesis that the standard deviation is less than 0.16 oz. The calculated P-value is approximately 0.9863.
Step 4 ：The final answer is: The correct null and alternative hypotheses are \(H_{0}: \sigma=0.16 \mathrm{oz}\) and \(H_{1}: \sigma<0.16 \mathrm{oz}\). The test statistic is \(\boxed{11.344}\). The P-value is \(\boxed{0.9863}\).
