Save Workers at a certain soda drink factory collected data on the volumes (in ounces) of a simple random sample of 20 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 $12.03 \mathrm{oz}$ and $12.59 \mathrm{oz}$, the range rule of thumb can be used to estimate that the standard deviation should be less than $0.14 \mathrm{oz}$. Use the sample data to test the claim that the population of volumes has a standard deviation less than $0.14 \mathrm{oz}$. Use a 0.05 significance level. Complete parts (a) through (d) below. a. Identify the null and alternative hypotheses. Choose the correct answer below. A. \[ \begin{array}{l} H_{0}: \sigma=0.14 \mathrm{oz} \\ H_{1}: \sigma \neq 0.14 \mathrm{oz} \end{array} \] C. \[ \begin{array}{l} H_{0}: \sigma \geq 0.14 \mathrm{oz} \\ H_{1}: \sigma<0.14 \mathrm{oz} \end{array} \] b. Compute the test statistic. \[ \chi^{2}=\square \] (Round to three decimal places as needed.) B. \[ \begin{array}{l} H_{0}: \sigma=0.14 \mathrm{oz} \\ H_{1}: \sigma<0.14 \mathrm{oz} \end{array} \] D. \[ \begin{array}{l} H_{0}: \sigma>0.14 \mathrm{oz} \\ H_{1}: \sigma=0.14 \mathrm{oz} \end{array} \]
Solution
Step 1 ：The null hypothesis (H0) is that the standard deviation is equal to 0.14 oz, and the alternative hypothesis (H1) is that the standard deviation is less than 0.14 oz.
Step 2 ：The test statistic can be computed using the formula for the chi-square statistic: \(\chi^{2} = \frac{(n-1)s^{2}}{\sigma^{2}}\), where n is the sample size, s is the sample standard deviation, and \(\sigma\) is the standard deviation under the null hypothesis.
Step 3 ：Substitute the given values into the formula: \(\chi^{2} = \frac{(20-1)(0.11)^{2}}{(0.14)^{2}}\)
Step 4 ：Simplify the equation: \(\chi^{2} = \frac{19*0.0121}{0.0196}\)
Step 5 ：Calculate the test statistic: \(\chi^{2} = 11.76\)
Step 6 ：So, the test statistic is \(\boxed{11.76}\) (rounded to two decimal places).
