Question

use a 0.025 significance level to test the claim that peanut candies have weights that vary more than plain candies. the standard deviation for the weights of plain candies is 0.331. a sample of 41 peanut candies has weights with a standard deviation of 0.28. assume the population is normally distributed. (use a comma to separate answers as needed. round to three decimal places as needed.) identify the rejection region(s). a. b. c. (c) use the \\(\chi^2\\)-test to find the standardized test statistic. 28.623 (round to three decimal places as needed.) (d) decide whether to reject or fail to reject the null hypothesis. reject fail to reject

Explanation:

Step1: Recall the chi - square test for variance

The test statistic for a chi - square test of variance is given by the formula \(\chi^{2}=\frac{(n - 1)s^{2}}{\sigma^{2}}\), where \(n\) is the sample size, \(s\) is the sample standard deviation, and \(\sigma\) is the population standard deviation. Here, \(n = 41\), \(s=0.28\), \(\sigma = 0.331\).
We calculated the test statistic \(\chi^{2}=\frac{(41 - 1)\times(0.28)^{2}}{(0.331)^{2}}=\frac{40\times0.0784}{0.109561}=\frac{3.136}{0.109561}\approx28.623\) (which matches the given value).

Step2: Determine the degrees of freedom and critical value

The degrees of freedom \(df=n - 1=41- 1 = 40\). For a one - tailed test (since we are testing if the variance of peanut candies is more than that of plain candies) with \(\alpha = 0.025\), we look up the critical value of \(\chi^{2}\) with \(df = 40\) and \(\alpha=0.025\). From the chi - square distribution table, \(\chi_{0.025,40}^{2}\approx59.342\).

Step3: Compare the test statistic with the critical value

Our test statistic \(\chi^{2}=28.623\) and the critical value \(\chi_{0.025,40}^{2}=59.342\). Since \(28.623<59.342\), the test statistic does not fall in the rejection region. So we fail to reject the null hypothesis.

Answer:

Fail to reject