Test the given claim. Assume that a simple random sample is selected from a normally distributed population. Use either the P-value method or the traditional method of testing hypotheses. Company A uses a new production method to manufacture aircraft altimeters. A simple random sample of new altimeters resulted in errors listed below. Use a 0.05 level of significance to test the claim that the new production method has errors with a standard deviation greater than $32.2 \mathrm{ft}$, which was the standard deviation for the old production method. If it appears that the standard deviation is greater, does the new production method appear to be better or worse than the old method? Should the company take any action? \[ \begin{array}{l} -44,78,-25,-75,-44,14,17,52,49,-50,-108, \\ -108 \end{array} \] What are the null and alternative hypotheses? A. \[ \begin{array}{r} H_{0}: \sigma=32.2 \mathrm{ft} \\ H_{1}: \sigma>32.2 \mathrm{ft} \\ \text { C. } \mathrm{H}_{0}: \sigma \neq 32.2 \mathrm{ft} \\ \mathrm{H}_{1}: \sigma=32.2 \mathrm{ft} \\ \text { E. } \mathrm{H}_{0}: \sigma<32.2 \mathrm{ft} \\ \mathrm{H}_{1}: \sigma=32.2 \mathrm{ft} \end{array} \] B. \[ \begin{array}{l} \mathrm{H}_{0}: \sigma=32.2 \mathrm{ft} \\ \mathrm{H}_{1}: \sigma \neq 32.2 \mathrm{ft} \end{array} \] D. \[ \text { D. } \begin{aligned} H_{0}: \sigma>32.2 \mathrm{ft} \\ H_{1}: \sigma=32.2 \mathrm{ft} \\ \text { F. } H_{0}: \sigma=32.2 \mathrm{ft} \\ H_{1}: \sigma<32.2 \mathrm{ft} \end{aligned} \] Find the test statistic. \[ \chi^{2}= \] (Round to two decimal places as needed.)
Solution
Step 1 ：State the null hypothesis and the alternative hypothesis. The null hypothesis is that the standard deviation is equal to \(32.2 ft\), and the alternative hypothesis is that the standard deviation is greater than \(32.2 ft\). So, the null and alternative hypotheses are: \[H_{0}: \sigma=32.2 ft\] \[H_{1}: \sigma>32.2 ft\]
Step 2 ：Calculate the test statistic. The test statistic is a measure of how far our data is from the null hypothesis. Since we are dealing with standard deviations, we will use the chi-square test statistic. The formula for the chi-square test statistic is: \[\chi^{2} = \frac{(n-1)s^{2}}{\sigma^{2}}\] where \(n\) is the sample size, \(s\) is the sample standard deviation, and \(\sigma\) is the population standard deviation.
Step 3 ：Substitute the given values into the formula. The sample size \(n\) is 12, the sample standard deviation \(s\) is approximately 62.28, and the population standard deviation \(\sigma\) is 32.2. So, \[\chi^{2} = \frac{(12-1)(62.28)^{2}}{(32.2)^{2}}\]
Step 4 ：Solve the equation to find the test statistic. The test statistic is approximately 41.15.
Step 5 ：The final answer is: The null and alternative hypotheses are: \[H_{0}: \sigma=32.2 ft\] \[H_{1}: \sigma>32.2 ft\] The test statistic is \(\boxed{41.15}\).
