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.

(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.
\\(\circ\\) reject
\\(\checkmark\\) fail to reject
(e) interpret the decision in the context of the original claim.
does it appear that peanut candies have weights that vary more than those of plain candies?
\\(\circ\\) a. since the null hypothesis is rejected, the weights appear to vary more for peanut candies. \\(\circ\\) b. since the null hypothesis is rejected, the weights do not appear to vary more for peanut candies.
\\(\circ\\) c. since the null hypothesis is not rejected, the weights do not appear to vary more for peanut candies. \\(\circ\\) d. since the null hypothesis is not rejected, the weights appear to vary more for peanut candies.

Explanation:

Part (c)

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

The formula for the chi - square test statistic for 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 $n = 41$, $s=0.28$, 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

$\chi^{2}=\frac{40\times0.0784}{0.109561}=\frac{3.136}{0.109561}\approx28.623$

Part (d)

To decide whether to reject or fail to reject the null hypothesis, we need to compare the test statistic with the critical value. For a right - tailed test with $\alpha = 0.025$ and $df=n - 1=40$, the critical value of $\chi^{2}$ (from the chi - square distribution table) is $\chi_{0.025,40}^{2}\approx55.758$. Our test statistic $\chi^{2}=28.623$ is less than the critical value $55.758$. So we fail to reject the null hypothesis.

Part (e)

The null hypothesis in this test is $H_{0}:\sigma\leq0.331$ (or $H_{0}:\sigma^{2}\leq0.109561$), and the alternative hypothesis (based on the claim that peanut candies have weights that vary more than plain candies) is $H_{1}:\sigma > 0.331$ (a right - tailed test). Since we failed to reject the null hypothesis, we do not have enough evidence to support the claim that the standard deviation (and thus the variance) of the weights of peanut candies is greater than that of plain candies. So the weights do not appear to vary more for peanut candies.

Answer:

28.623