Question

Testing Claims About Variation. In Exercises 5–16, test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. Assume that a simple random sample is selected from a normally distributed population.

Aircraft Altimeters The Skytel Avionics company uses a new production method to manufacture aircraft altimeters. A simple random sample of new altimeters resulted in the 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?

-42 78 -22 -72 -45 15 17 51 -5 -53 -9 -109

Short answer

The hypotheses are as follows.

\(\begin{array}{l}{H_0}:\sigma = 32.2\,\,\\{H_1}:\sigma > 32.2\end{array}\)

The test statistic is 29.176, and the critical value is 19.675. The hypothesis is rejected to conclude that there is sufficient evidence to support the claim.

The company should take some action because the variation is to be greater than the old production method. So, the new method appears to be worse.

Step-by-step solution

Given information

The new method of manufacturing altimeters results in the following errors:

-42 78 -22 -72 -45 15 17 51 -5 -53 -9 -109

The level of significance is 0.05.

The claim states that the standard deviation of errors from the new method is greater than 32.2 ft, which is the measure for the standard deviation from the old method.

Describe the hypothesis

For applying the hypothesis test, first set up a null and an alternative hypothesis.

The null hypothesis is the statement about the value of a population parameter, which is equal to the claimed value. It is denoted by\({H_0}\).

The alternate hypothesis is a statement that the parameter has a value that is opposite to the null hypothesis. It is denoted by\({H_1}\).

State the null and alternative hypotheses

Let\(\sigma \)be the actual standard deviation for the errors from the new method of manufacturing altimeters.

From the claim, the null and alternative hypotheses are as follows.

\(\begin{array}{l}{H_0}:\sigma = 32.2\,\,\\{H_1}:\sigma > 32.2\end{array}\)

Find the sample standard deviation

LetX be the simple random sample of errors. From the new method of manufacturing altimeters, it is calculated as follows.

-42 78 -22 -72 -45 15 17 51 -5 -53 -9 -109

The sample mean of X is computed as follows.

\(\begin{array}{c}\bar x = \frac{{\sum\limits_{i = 1}^n {{x_i}} }}{{\rm{n}}}\\ = \frac{{ - 42 + 78 + ... + \left( { - 109} \right)}}{{12}}\\ = - 16.3333\end{array}\)

The sample standard deviation is calculated as follows.

\(\begin{array}{c}s = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{\left( {{x_i} - \bar x} \right)}^2}} }}{{n - 1}}} \\ = \sqrt {\frac{{{{\left( { - 42 - \left( { - 16.3333} \right)} \right)}^2} + {{\left( {78 - \left( { - 16.3333} \right)} \right)}^2} + ... + {{\left( { - 109 - \left( {16.3333} \right)} \right)}^2}}}{{12 - 1}}} \\ = 52.4410\end{array}\)

Thus, the sample standard deviation is 52.4410.

Compute the test statistic

To conduct a hypothesis test of a claim about a population standard deviation\(\sigma \) or population variance\({\sigma ^2}\),the test statistic is computed as follows.

\(\begin{array}{c}{\chi ^2} = \frac{{\left( {{\rm{n}} - 1} \right) \times {s^2}}}{{{\sigma ^2}}}\\ = \frac{{\left( {12 - 1} \right) \times {{52.4410}^2}}}{{{{32.2}^2}}}\\ = 29.176\end{array}\).

Thus, the value of the test statistic is 29.176.

The degree of freedom is as follows.

\(\begin{array}{c}df = n - 1\\ = 12 - 1\\ = 11\end{array}\)

Find the critical value

The critical value\(\chi _{0.05}^2\)is obtained using the chi-square table, as follows.

\(\begin{array}{c}P\left( {{\chi ^2} > \chi _\alpha ^2} \right) = \alpha \\P\left( {{\chi ^2} > \chi _{0.05}^2} \right) = 0.05\end{array}\)

Refer to the chi-square table for the critical value of 19.675, corresponding to the area of 0.05 and the degree of freedom 11.

 Step 7: State the decision

The decision rule for the test is as follows.

If\({\chi ^2} > \chi _{0.05}^2\),reject the null hypothesis at a given level of significance. Otherwise, fail to reject the null hypothesis.

As it is observed that \({\chi ^2} = 29.176\, > \,\chi _{0.05}^2 = 19.675\), the null hypothesis is rejected.

Conclusion

Thus, there is enough evidence to supportthe 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.

The variation appears to be greater than that in the old production method. So, the new method appears to be worse.

The company should take immediate action to reduce the variation.