The image contains a problem statement for testing a statistical claim using hypothesis testing. Here’s a detailed transcription and explanation suitable for an educational website:---**Problem Statement:**A simple random sample from a normally distributed population is selected. Use either the P-value method or the traditional method of testing hypotheses.**Scenario:**Company A employs a new production method to manufacture aircraft altimeters, and a simple random sample of new altimeters yielded errors as follows:- Errors: -43, 79, -25, -75, -42, 10, 16, 53, -9, -52, -106, -106**Objective:**Using a significance level of 0.05, test if the new production method results in errors with a standard deviation greater than 32.2 ft. The standard deviation for the old production method was 32.2 ft. Decide if the new production method is better or worse and if the company should take action.**Hypotheses:**Select the appropriate null and alternative hypotheses:- (A) \( H_0: \sigma = 32.2 \) ft, \( H_1: \sigma > 32.2 \) ft- (B) \( H_0: \sigma = 32.2 \) ft, \( H_1: \sigma < 32.2 \) ft- (C) \( H_0: \sigma = 32.2 \) ft, \( H_1: \sigma \neq 32.2 \) ft- (D) \( H_0: \overline{x} = 32.2 \) ft, \( H_1: \overline{x} > 32.2 \) ft- (E) \( H_0: \overline{x} = 32.2 \) ft, \( H_1: \overline{x} < 32.2 \) ft- (F) \( H_0: \overline{x} = 32.2 \) ft, \( H_1: \overline{x} \neq 32.2 \) ftChoose option (A), testing if the standard deviation is greater than 32.2 ft.**Test Statistic:**Calculate the test statistic \(\chi^2 = 36.62\).**Critical Value:**Determine the critical value(s):

Name: The image contains a problem statement for testing a statistical clai
Uploaded: 2024-03-20T07:07:12.000Z
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Description: The image contains a problem statement for testing a statistical claim using hypothesis testing. Here’s a detailed transcription and explanation suitable for an educational website:---**Problem Statement:**A simple random sample from a normally dist

The standard deviation of production errors is calculated by using excel as follows:The null and alternative hypotheses are given as follows: The sample standard deviation is calculated as, s=58.017. The sample size, n is 12. Hence, the degrees of freedom is, (n-1)=12–1=11. The test statistic is given and calculated as follows:Computation of P-value: The right-tailed P-value for the chi-square distribution at 11 degrees of freedom is obtained using the excel formula, “=CHISQ.DIST.RT(35.71,11)”. Decision rule: If p-value ≤α, then reject the null hypothesis. Otherwise, do not reject the null hypothesis. Conclusion: Here, the p-value (approximately 0) is less than the level of significance (0.05). Therefore, reject the null hypothesis. There is sufficient evidence to conclude that the standard deviation of error is more than 32.2. The variation appears to be more than the past, so the new method appears to be less effective because there will be more altimeters that have errors. Therefore, the company should take immediate action to reduce the variation.