Bags of a certain brand of tortilla chips claim to have a net weight of 14 oz.Net weights slightly vary from bag to bag.Assume net weights are Normally distributed.A representative of a consumer advocate group wishes to see if there is any evidence that the mean net weight is less than advertised and so intends to test the hypotheses H 0 :  = 14 versus H a : 

Question

Bags of a certain brand of tortilla chips claim to have a net weight of 14 oz.Net weights slightly vary from bag to bag.Assume net weights are Normally distributed.A representative of a consumer advocate group wishes to see if there is any evidence that the mean net weight is less than advertised and so intends to test the hypotheses H₀:  = 14 versus H_a:  < 14.To do this,he selects 16 bags of tortilla chips of this brand at random and determines the net weight of each.He finds a sample mean of 13.88 oz with a standard deviation of s = 0.24 oz.Assume the standard deviation for the distribution of actual net weights for bags of tortilla chips of this brand is  = 0.25.At the 5% significance level,what is the power of our test when,in fact, = 13.8 oz?

A) 0.654
B) 0.857
C) 0.940
D) 0.997

Miguel Cuautle · Accepted Answer

Final Answer : C, Explanation :  The power of the test is the probability of rejecting the null hypothesis when it is false. In this case, the null hypothesis is that the mean net weight is 14 oz, and the alternative hypothesis is that the mean net weight is less than 14 oz. The power of the test is the probability of rejecting the null hypothesis when the mean net weight is actually 13.8 oz.To calculate the power of the test, we need to find the critical value for the test statistic. The critical value is the value of the test statistic that separates the rejection region from the non-rejection region. In this case, the test statistic is the sample mean, and the critical value is the value of the sample mean that would lead to a p-value of 0.05.The p-value is the probability of getting a sample mean as extreme as the one we observed, assuming that the null hypothesis is true. In this case, the p-value is the probability of getting a sample mean of 13.88 oz or less, assuming that the mean net weight is 14 oz.The power of the test is the probability of rejecting the null hypothesis when the mean net weight is actually 13.8 oz. This is the same as the probability of getting a sample mean of 13.88 oz or less, assuming that the mean net weight is 13.8 oz.To calculate the power of the test, we need to find the probability of getting a sample mean of 13.88 oz or less, assuming that the mean net weight is 13.8 oz. This is the same as the probability of getting a z-score of -2.88 or less, where the z-score is the difference between the sample mean and the mean net weight, divided by the standard deviation.The probability of getting a z-score of -2.88 or less is 0.002. Therefore, the power of the test is 0.002.

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