Bags of a certain brand of tortilla chips claim to have a net weight of 14 ounces.Net weights actually vary slightly 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 ounces.Net weights actually vary slightly 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 ounces with a standard deviation of s = 0.24 ounces.Determine which of the following statements is true,regarding the decision the representative would make.

A) He would not reject H₀ at a significance level of 0.05.
B) He would reject H₀ at a significance level of 0.05 but not at 0.025.
C) He would reject H₀ at a significance level of 0.025 but not at 0.01.
D) He would reject H₀ at a significance level of 0.01.

Alicia Aurelio · Accepted Answer

Final Answer : B, Explanation :  We need to conduct a one-sided hypothesis test with a significance level of 0.05.The null hypothesis is that the mean net weight is equal to 14 ounces: H0: μ = 14The alternative hypothesis is that the mean net weight is less than 14 ounces: Ha: μ < 14Given that the sample size n = 16 and the population standard deviation is unknown, we will use a t-test. The test statistic is:t = (x̄ - μ) / (s / sqrt(n)) = (13.88 - 14) / (0.24 / sqrt(16)) = -2.67The degrees of freedom is n - 1 = 15. Using a t-table, the critical value for a one-sided test with a significance level of 0.05 and 15 degrees of freedom is -1.753. Since our calculated t-value (-2.67) is less than the critical value (-1.753), we reject the null hypothesis.At a significance level of 0.025, the critical value is -2.131. Since -2.67 is less than -2.131, we would still reject the null hypothesis.At a significance level of 0.01, the critical value is -2.602. Since -2.67 is less than -2.602, we would reject the null hypothesis.Therefore, the best choice is B because we reject the null hypothesis at a significance level of 0.05 but not at 0.025.

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