**Hypothesis Testing for Aircraft Altimeter Errors****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.**Scenario:** 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 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?**Error Data Sample:** -44, 76, -25, -70, -40, 11, 19, 54, -54, -108, -108**Steps:**1. **Find the test statistic:** \[ \chi^2 = \_\_ \] (Round to two decimal places as needed.)2. **Determine the critical value(s):** \[ \text{The critical value(s) is/are } \_\_ \] (Use a comma to separate answers as needed. Round to two decimal places as needed.)3. **Decision Rule:** - Since the test statistic is **greater than** the critical value(s), - **Reject \( H_0 \):** There is **sufficient** evidence to support the claim that the new production method has errors with a standard deviation greater than 32.2 ft. - The variation appears to be **greater** than in the past, so the new method appears to be **worse** because there will be larger errors.### Diagrams and Graphs- **Explanation:** No explicit diagrams or graphs are present in the image. The focus is on calculating the test statistic and determining critical values for hypothesis testing based on provided data.The company should consider revisiting the production method to reduce errors.

The formula for the Chi-Square test for single population variance is as follows: χ2 = (n-1)s2σ2 The Excel formula for the calculation of the critical value of Chi-Square is as follows: =CHIIV(probability , deg_freedom) From the information, observe that new production method has errors with standard deviation is greater than 32.2 ft. Therefore, the Null and Alternative hypothesis is as follows: Standard deviation is 32.2 and standard deviation greater than 32.2 ft. The level of significance is 0.05 The sample size is 12. The sample standard deviation is 62.13 Therefore, observation xi (xi-mean) (x-mean)2 1 -44 1.230769231 1.514792899 2 76 121.2307692 14696.89941 3 -25 20.23076923 409.2840237 4 -70 -24.7692308 613.5147929 5 -40 5.230769231 27.36094675 6 11 56.23076923 3161.899408 7 19 64.23076923 4125.591716 8 54 99.23076923 9846.745562 9 -5 40.23076923 1618.514793 10 -54 -8.76923077 76.89940828 11 -108 -62.7692308 3939.976331 12 -108 -62.7692308 3939.976331 Sum -294 42458.17751 Average -45.2308 3859.83432 squre root 62.12756489 Now, value of the test statistics is: χ2 = (n-1)s2σ2 = (12-1)*(62.13)2(32.2)2= (11)*(62.13)2(32.2)2= 42461.50591036.84=42461.50591036.84=40.9528 χ2 = 40.95 The degrees of freedom is, d.f = n-1 = 12-1d.f =11 Therefore, the critical value is, χ2 = 19.675 (=CHIINV(0.05,11)) From the calculated value of the statistic and critical value, observe that the Null hypothesis is rejected. Since the statistic is greater than the critical values are reject H0. There is sufficient evidence to support the claim that the the new production method has error with standard deviation is greater than 32.2 Therefore, the company should take immediate action to reduce the variation.