A manufacturer wants to increase the shelf life of a line of cake mixes. Past records indicate that the average shelf life of the mix is 216 days. After a revised mix has been developed, a sample of nine boxes of cake mix gave these shelf lives (in days): 215, 217, 218, 219, 216, 217, 217, 218, and 218. Using α = 0.025, has the shelf life of the cake mix increased?a. Yes, because computed t is less than the critical value.b. Yes, because computed t is greater than the critical value.c. No, because computed t lies in the region of acceptance.d. No, because 217.24 is quite close to 216.
Accepted Solution
A:
Answer:Option b.Step-by-step explanation:Sample size n = 9sample mean = 217.222Standard deviation s = 1.202alpha a = 0.025H0 : u = 216H1 : u > 216 ( claim)Now run a T test and result is :Test statistic, t = [tex]\frac{(xbar-u)}{\frac{s}{\sqrt{n}}}[/tex] t = [tex]\frac{(217.222-216)}{\frac{1.202}{\sqrt{9}}}[/tex] = [tex]\frac{1.222}{0.40066}}[/tex] t = 3.0509Critical value = t(a,n-1) = t(0.025,9-1) = 2.306Since t > critical value, Hence reject H0.Option b. is the answer. Yes, because computed t is greater than the critical value.