Question

quarters are currently minted with weights normally distributed and having a standard deviation of 0.066. new equipment is being tested in an attempt to improve quality by reducing variation. a simple random sample of 28 quarters is obtained from those manufactured with the new equipment, and this sample has a standard deviation of 0.044. use a 0.05 significance level to test the claim that quarters manufactured with the new equipment have weights with a standard deviation less than 0.066. does the new equipment appear to be effective in reducing the variation of weights? (c) use the \\(\chi^2\\)-test to find the standardized test statistic. \\(\chi^2 = \\) 12 (round to three decimal places as needed.) (d) decide whether to reject or fail to reject the null hypothesis. \\(\circ\\) fail to reject \\(\checkmark\\) reject (e) interpret the decision in the context of the original claim. \\(\circ\\) a. since the null hypothesis is rejected, the new equipment appears to be more effective. \\(\circ\\) b. since the null hypothesis is not rejected, the new equipment appears to be more effective. \\(\circ\\) c. since the null hypothesis is rejected, the new equipment does not appear to be more effective. \\(\circ\\) d. since the null hypothesis is not rejected, the new equipment does not appear to be more effective.

Explanation:

(c)

Step1: Recall the formula for the chi - square test statistic for variance

The formula for the chi - square test statistic when testing a claim about a population variance (or standard deviation) 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.

Step2: Identify the values

We are given that \(n = 28\) (so \(n - 1=27\)), \(s=0.044\), and \(\sigma = 0.066\)

Step3: Calculate \(s^{2}\) and \(\sigma^{2}\)

\(s^{2}=(0.044)^{2}=0.001936\) and \(\sigma^{2}=(0.066)^{2}=0.004356\)

Step4: Substitute the values into the formula

\(\chi^{2}=\frac{(28 - 1)\times0.001936}{0.004356}=\frac{27\times0.001936}{0.004356}\)

First, calculate the numerator: \(27\times0.001936 = 0.052272\)

Then, divide the numerator by the denominator: \(\frac{0.052272}{0.004356}=12\)

(d)

Step1: Determine the critical value

For a left - tailed test with \(\alpha = 0.05\) and \(df=n - 1=27\), we look up the critical value of \(\chi^{2}\) in the chi - square distribution table. The critical value \(\chi_{0.95,27}^{2}\) (since it's a left - tailed test, we use \(\alpha = 0.05\) in the left tail) is approximately 16.151 (wait, no. Wait, for a left - tailed test, the critical value is \(\chi_{1-\alpha,df}^{2}\). For \(\alpha = 0.05\) and \(df = 27\), \(\chi_{0.95,27}^{2}\approx16.151\)? Wait, no, I think I mixed up. Wait, the test statistic we calculated is \(\chi^{2}=12\), and for a left - tailed test, we reject \(H_{0}\) if \(\chi^{2}<\chi_{1 - \alpha,df}^{2}\)

Wait, the null hypothesis \(H_{0}:\sigma = 0.066\) and the alternative hypothesis \(H_{1}:\sigma<0.066\)

The degrees of freedom \(df=n - 1 = 27\)

Looking up the chi - square table, \(\chi_{0.95,27}^{2}\) (the critical value for a left - tailed test with \(\alpha = 0.05\)): The chi - square distribution is right - skewed. For a left - tailed test, the critical value is such that \(P(\chi^{2}<\chi_{1-\alpha,df}^{2})=\alpha\)

From the chi - square table, for \(df = 27\) and \(\alpha=0.05\) (left - tailed), \(\chi_{0.95,27}^{2}\approx16.151\)? Wait, no, actually, when we calculated the test statistic \(\chi^{2}=12\), and since \(12<\chi_{0.95,27}^{2}\) (let's check a more accurate table. The critical value for \(df = 27\) and \(\alpha = 0.05\) (left - tailed) is approximately 16.151? Wait, no, I think I made a mistake. Wait, the test statistic is \(\chi^{2}=\frac{(n - 1)s^{2}}{\sigma^{2}}\). In a left - tailed test, we reject \(H_{0}\) when the test statistic is less than the critical value.

The critical value for \(df = 27\) and \(\alpha=0.05\) (left - tailed) can be found using a chi - square calculator or table. The critical value \(\chi_{0.95,27}^{2}\) is approximately 16.151? Wait, no, actually, the correct critical value for \(df = 27\) and \(\alpha = 0.05\) (left - tailed) is \(\chi_{0.95,27}^{2}=16.151\)? Wait, no, let's use the formula or a calculator. Alternatively, since our test statistic is 12, and the critical value for a left - tailed test with \(\alpha = 0.05\) and \(df = 27\) is \(\chi_{0.95,27}^{2}\approx16.151\). Since \(12<16.151\), we reject the null hypothesis.

(e)

The original claim is that the new equipment has a standard deviation less than 0.066 (i.e., it reduces variation). The null hypothesis \(H_{0}:\sigma = 0.066\) and the alternative hypothesis \(H_{1}:\sigma<0.066\). When we reject the null hypothesis, it means that there is sufficient evidence to support the alternative hypothesis. So, since we rejected \(H_{0}\), we can conclude that the new equipment appears to be more effective in reducing the variation of weights.

Answer:

11.643 (Let's recalculate properly to confirm. The formula for the chi - square test statistic for a test of a single variance 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.

Here, \(n = 28\), so \(n-1=27\), \(s = 0.044\), and \(\sigma=0.066\)

First, calculate \(s^{2}=(0.044)^{2}=0.001936\) and \(\sigma^{2}=(0.066)^{2}=0.004356\)

Then \(\chi^{2}=\frac{27\times0.001936}{0.004356}=\frac{0.052272}{0.004356}\approx11.999\)? Wait, maybe I made a mistake earlier. Wait, \(27\times0.001936 = 27\times1.936\times10^{- 3}=52.272\times10^{-3}=0.052272\)

\(0.052272\div0.004356 = 12\)? Wait, maybe the initial answer of 12 is a rounded value. Let's do the division: \(0.052272\div0.004356=\frac{52.272}{4.356}=12\) (exactly? Let's check \(4.356\times12 = 52.272\). Yes, so the correct value is 12.000 when rounded to three decimal places? Wait, maybe my first calculation was wrong. Let's re - express:

\(s = 0.044\), so \(s^{2}=0.044\times0.044 = 0.001936\)

\(n - 1=27\)

\(\sigma^{2}=0.066\times0.066 = 0.004356\)

\(\chi^{2}=\frac{(28 - 1)\times(0.044)^{2}}{(0.066)^{2}}=\frac{27\times0.001936}{0.004356}=\frac{0.052272}{0.004356}=12\) (exactly, because \(0.004356\times12 = 0.052272\)) So the answer is 12.000 (or 12 when rounded to three decimal places)