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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.

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 $H_1$.

Let $蟽$ 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.

$$\begin{gathered} H_0:蟽=32.2 \\ {H}_1:蟽>32.2 \end{gathered}$$

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.

$$\begin{gathered} \stackrel{炉}{x}=\frac{{鈭�}_{i=1}^n{x}_i}{\text{n}} \\ =\frac{-42+78+...+\left(-109\right)}{12} \\ =-16.3333 \end{gathered}$$

The sample standard deviation is calculated as follows.

$$\begin{gathered} s=\sqrt{\frac{{鈭�}_{i=1}^n{\left(x_i-\stackrel{炉}{x}\right)}^2}{n-1}} \\ =\sqrt{\frac{{\left(-42-\left(-16.3333\right)\right)}^2+{\left(78-\left(-16.3333\right)\right)}^2+...+{\left(-109-\left(16.3333\right)\right)}^2}{12-1}} \\ =52.4410 \end{gathered}$$

Thus, the sample standard deviation is 52.4410.

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

$$\begin{gathered} {蠂}^2=\frac{\left(\text{n}-1\right)脳s^2}{{蟽}^2} \\ =\frac{\left(12-1\right)脳{52.4410}^2}{{32.2}^2} \\ =29.176 \end{gathered}$$

Thus, the value of the test statistic is 29.176.

The degree of freedom is as follows.

$$\begin{gathered} df=n-1 \\ =12-1 \\ =11 \end{gathered}$$

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

$$\begin{gathered} P\left({蠂}^2>{蠂}_{伪}^2\right)=伪 \\ P\left({蠂}^2>{蠂}_{0.05}^2\right)=0.05 \end{gathered}$$

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.

The decision rule for the test is as follows.

If ${蠂}^2>{蠂}_{0.05}^2$, reject the null hypothesis at a given level of significance. Otherwise, fail to reject the null hypothesis.

As it is observed that ${蠂}^2=29.176鈥�>鈥�{蠂}_{0.05}^2=19.675$, the null hypothesis is rejected.

Thus, there is enough evidence to support 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.

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.