Question

a snack food manufacturer estimates that the variance of the number of grams of carbohydrates in chips is 1.37. a dietician is asked to test this claim and finds that a random sample of 19 servings h 1.36. at α = 0.10, is there enough evidence to reject the manufacturer’s claim? assume the populat distributed. complete parts (a) through (e) below. the critical value(s) is(are) . (round to two decimal places as needed. use a comma to separate answers as needed.) choose the correct statement below and fill in the corresponding answer boxes. a. the rejection region is $\chi^2 > $ . b. the rejection region is $\chi^2 < $ . c. the rejection regions are $\chi^2 < $ and $\chi^2 > $ . (c) find the standardized test statistic $\chi^2$. (round to two decimal places as needed.) (d) decide whether to reject or fail to reject the null hypothesis, and (e) interpret the decision in th claim. fill in the correct answers below.

Explanation:

Step1: Identify the test type

This is a chi - square test for variance. The sample size \(n = 19\), so the degrees of freedom \(df=n - 1=19 - 1 = 18\). The significance level \(\alpha=0.10\). Since we are testing a claim about variance (a two - tailed test, because we are just testing if there is enough evidence to reject the claim, not a one - tailed direction), we will have two critical values.

Step2: Find the critical values

For a two - tailed chi - square test with \(\alpha = 0.10\) and \(df = 18\), we find the lower critical value \(\chi_{1-\alpha/2}^{2}\) and the upper critical value \(\chi_{\alpha/2}^{2}\).
\(\alpha/2=0.05\) and \(1-\alpha/2 = 0.95\)
Using the chi - square distribution table or a calculator, \(\chi_{0.95}^{2}(18)=9.390\) and \(\chi_{0.05}^{2}(18)=28.869\) (rounded to three decimal places, for two decimal places: \(\chi_{0.95}^{2}(18)\approx9.39\), \(\chi_{0.05}^{2}(18)\approx28.87\))

Step3: Determine the rejection region

Since it is a two - tailed test, the rejection regions are \(\chi^{2}<\chi_{1 - \alpha/2}^{2}\) and \(\chi^{2}>\chi_{\alpha/2}^{2}\), so option C is correct.

Step4: Calculate the test statistic

The formula for the chi - square test statistic for variance is \(\chi^{2}=\frac{(n - 1)s^{2}}{\sigma^{2}}\), where \(s^{2}\) is the sample variance and \(\sigma^{2}\) is the population variance.
We are given that \(\sigma^{2}=1.37\) (the manufacturer's claim) and \(s^{2}=1.36\), \(n = 19\)
\(\chi^{2}=\frac{(19 - 1)\times1.36}{1.37}=\frac{18\times1.36}{1.37}=\frac{24.48}{1.37}\approx17.87\) (rounded to two decimal places)

Step5: Decision about the null hypothesis

We compare the test statistic \(\chi^{2}=17.87\) with the critical values. Since \(9.39<17.87<28.87\), we fail to reject the null hypothesis.

For the critical value part:

The critical values are approximately \(9.39\) and \(28.87\). The rejection region is option C with \(\chi^{2}<9.39\) and \(\chi^{2}>28.87\)

For the test statistic part:

The standardized test statistic \(\chi^{2}\approx17.87\)

For the decision part:

Since the test statistic \(\chi^{2}=17.87\) is between the lower critical value \(9.39\) and the upper critical value \(28.87\), we fail to reject the null hypothesis. This means there is not enough evidence to reject the manufacturer's claim at the \(\alpha = 0.10\) significance level.

Answer:

• Critical values: \(9.39, 28.87\)
• Rejection region: Option C, with \(\chi^{2}<9.39\) and \(\chi^{2}>28.87\)
• Test statistic: \(\approx17.87\)
• Decision: Fail to reject the null hypothesis. There is not enough evidence to reject the manufacturer's claim.