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.

(b) find the critical value(s).
59.342
(use a comma to separate answers as needed. round to three decimal places as needed.)
identify the rejection region(s).

(c) use the \\(\chi^2\\)-test to find the standardized test statistic.
(round to three decimal places as needed.)

Explanation:

Step1: Recall the formula for the chi - square test statistic for standard deviation

The formula for the chi - square test statistic \(\chi^{2}=\frac{(n - 1)s^{2}}{\sigma^{2}}\), 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 \(n = 41\), \(s=0.28\), and \(\sigma = 0.331\).
First, calculate \((n - 1)\): \(n-1=41 - 1=40\).
Then, calculate \(s^{2}\): \(s^{2}=(0.28)^{2}=0.0784\).
And \(\sigma^{2}=(0.331)^{2}=0.109561\).

Step3: Substitute the values into the formula

\(\chi^{2}=\frac{40\times0.0784}{0.109561}\)
First, calculate the numerator: \(40\times0.0784 = 3.136\)
Then, divide the numerator by the denominator: \(\chi^{2}=\frac{3.136}{0.109561}\approx30.765\)

Answer:

30.765