Question

quarters are currently minted with weights normally distributed and having a standard deviation of 0.069. new equipment is being tested in an attempt to improve quality by reducing variation. a simple random sample of 30 quarters is obtained from those manufactured with the new equipment, and this sample has a standard deviation of 0.047. 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 0.069. does the new equipment appear to be effective in reducing the variation of weights?\\(\chi^2 = \square\\) (round to three decimal places as needed.)(d) decide whether to reject or fail to reject the null hypothesis.\\(\bigcirc\\) reject\\(\bigcirc\\) fail to reject(e) interpret the decision in the context of the original claim.\\(\bigcirc\\) a. since the null hypothesis is rejected, the new equipment does not appear to be more effective.\\(\bigcirc\\) b. since the null hypothesis is rejected, the new equipment appears to be more effective.\\(\bigcirc\\) c. since the null hypothesis is not rejected, the new equipment appears to be more effective.\\(\bigcirc\\) d. since the null hypothesis is not rejected, the new equipment does not appear to be more effective

Explanation:

Step1: Identify the test and formula

We are testing a claim about the standard deviation, so we use the chi - square test for variance (since standard deviation is related to variance). The formula for the chi - square test statistic when testing a claim about the standard deviation (or 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.

Given: $n = 30$, $s=0.047$, $\sigma = 0.069$.

Step2: Calculate the chi - square statistic

First, calculate $s^{2}$ and $\sigma^{2}$:
$s^{2}=(0.047)^{2}=0.002209$
$\sigma^{2}=(0.069)^{2}=0.004761$

Then, substitute into the formula:
$\chi^{2}=\frac{(30 - 1)\times0.002209}{0.004761}=\frac{29\times0.002209}{0.004761}$

Calculate $29\times0.002209 = 0.064061$

Then $\chi^{2}=\frac{0.064061}{0.004761}\approx13.455$

Step3: Determine the critical value and make a decision

The degrees of freedom $df=n - 1=30 - 1 = 29$.
We are conducting a left - tailed test with $\alpha = 0.05$. The critical value for $\chi^{2}$ with $df = 29$ and $\alpha=0.05$ (left - tailed) can be found from the chi - square distribution table. The critical value $\chi_{0.95,29}^{2}$ (since it's left - tailed, we use $1-\alpha$) is approximately 17.708.

Our test statistic $\chi^{2}=13.455$ is less than the critical value $17.708$. So we reject the null hypothesis.

Step4: Interpret the decision

The null hypothesis $H_{0}:\sigma = 0.069$ and the alternative hypothesis $H_{1}:\sigma<0.069$. Since we reject the null hypothesis, there is sufficient evidence to support the claim that the standard deviation is less than 0.069. So the new equipment appears to be more effective.

Answer:

For $\chi^{2}$: $\boldsymbol{13.455}$

(d) Reject

(e) B. Since the null hypothesis is rejected, the new equipment appears to be more effective.