Question 2 of 6, Step 3 of 3
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Claudia's bakery business has taken off. Her goods are so popular that she has decided to invest in kitchen equipment that will do some of the work for her. One machine in particular prepares bread dough, and it is set to measure 7 grams of yeast per loaf. Because baking requires precise measurements, she wants to test the new machine to make sure that the variance in the amounts of yeast per loaf is less than 0.27 , at the 0.10 tevel of significance. A sample of the amounts of yeast added to the dough for 13 loaves has a variance of 0.12 . Does this evidence support the claim that the new machine produces dough with a variance in the amounts of yeast per loaf of less than 0.27 ? Assume that the amounts of yeast per loaf are normally distributed.
Step 3 of 3 : Draw a conclusion and interpret the decision.
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We reject the null hypothesis and conclude that there is insufficient evidence at a 0.10 level of significance that the variance in the amounts of yeast per loaf is less than 0.27 .
We fail to reject the null hypothesis and conclude that there is insufficient evidence at a 0.10 level of significance that the variance in the amounts of yeast per loaf is less than 0.27 .
We reject the null hypothesis and conclude that there is sufficient evidence at a 0.10 level of significance that the variance in the amounts of yeast per loaf is less than 0.27 .
We fail to reject the null hypothesis and conclude that there is sufficient evidence at a 0.10 level of
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\(\boxed{\text{Final Answer: We fail to reject the null hypothesis and conclude that there is insufficient evidence at a 0.10 level of significance that the variance in the amounts of yeast per loaf is less than 0.27.}}\)
Step 1 ：The problem is asking us to determine whether there is sufficient evidence to support the claim that the variance in the amounts of yeast per loaf is less than 0.27. The null hypothesis is that the variance is equal to 0.27, and the alternative hypothesis is that the variance is less than 0.27.
Step 2 ：We are given a sample of 13 loaves, with a sample variance of 0.12. The hypothesized population variance is 0.27. We are also given a significance level of 0.10.
Step 3 ：We perform a hypothesis test for the variance. The test statistic is calculated as \( \chi^2 = \frac{(n-1)s^2}{\sigma^2} = \frac{(13-1)0.12}{0.27} = 5.33 \), where \( n = 13 \) is the sample size, \( s^2 = 0.12 \) is the sample variance, and \( \sigma^2 = 0.27 \) is the hypothesized population variance.
Step 4 ：The degrees of freedom for the test is \( df = n - 1 = 13 - 1 = 12 \).
Step 5 ：We find the p-value associated with the test statistic. The p-value is 0.9459, which is greater than the significance level of 0.10.
Step 6 ：Since the p-value is greater than the significance level, we fail to reject the null hypothesis. This means that there is insufficient evidence at a 0.10 level of significance to support the claim that the variance in the amounts of yeast per loaf is less than 0.27.
Step 7 ：\(\boxed{\text{Final Answer: We fail to reject the null hypothesis and conclude that there is insufficient evidence at a 0.10 level of significance that the variance in the amounts of yeast per loaf is less than 0.27.}}\)
