A potato chip manufacturer produces bags of potato chips that are supposed to have a net weight of 326 grams. Because the chips vary in size, it is difficult to fill the bags to the exact weight desired. However, the bags pass inspection so long as the standard deviation of their weights is no more than 3 grams. A quality control inspector wished to test the claim that one batch of bags has a standard deviation of more than 3 grams, and thus does not pass inspection. If a sample of 24 bags of potato chips is taken and the standard deviation is found to be 3.4 grams, does this evidence, at the 0.05 level of significance, support the claim that the bags should fail inspection? Assume that the weights of the bags of potato chips are normally distributed.
Step 3 of 3: Draw a conclusion and interpret the decision.
Answer
We reject the null hypothesis and conclude that there is sufficient evidence at a 0.05 level of significance that the bags should fail inspection.
We reject the null hypothesis and conclude that there is insufficient evidence at a 0.05 level of significance that the bags should fail inspection.
We fail to reject the null hypothesis and conclude that there is insufficient evidence at a 0.05 level of significance that the bags should fail inspection.
We fail to reject the null hypothesis and conclude that there is sufficient evidence at a 0.05 level of significance that the bags should fail inspection.
Answer
\(\boxed{\text{Final Answer: We fail to reject the null hypothesis and conclude that there is insufficient evidence at a 0.05 level of significance that the bags should fail inspection.}}\)
Step 1 ：The problem is asking us to perform a hypothesis test for the standard deviation. The null hypothesis is that the standard deviation is less than or equal to 3 grams, and the alternative hypothesis is that the standard deviation is more than 3 grams.
Step 2 ：We are given a sample size of 24, a sample standard deviation of 3.4 grams, and a significance level of 0.05.
Step 3 ：We need to calculate the test statistic and compare it to the critical value to decide whether to reject or fail to reject the null hypothesis.
Step 4 ：Given that the sample size (n) is 24, the sample standard deviation (s) is 3.4 grams, the population standard deviation (sigma) is 3 grams, and the significance level (alpha) is 0.05.
Step 5 ：The chi-square value is calculated as \(\chi^2 = \frac{(n-1)s^2}{\sigma^2} = \frac{(24-1)3.4^2}{3^2} = 29.54\)
Step 6 ：The critical value for a chi-square distribution with n-1 degrees of freedom at a significance level of 0.05 is 35.17.
Step 7 ：Since the chi-square value is less than the critical value, we fail to reject the null hypothesis.
Step 8 ：Thus, there is insufficient evidence at a 0.05 level of significance that the bags should fail inspection.
Step 9 ：\(\boxed{\text{Final Answer: We fail to reject the null hypothesis and conclude that there is insufficient evidence at a 0.05 level of significance that the bags should fail inspection.}}\)
