Question

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.

Answer 1

Answer:

Option b.

Step-by-step explanation:

Sample size n = 9

sample mean = 217.222

Standard deviation s = 1.202

alpha a = 0.025

H0 : u = 216

H1 : u > 216 ( claim)

Now run a T test and result is :

Test statistic, t = $\frac{(xbar-u)}{\frac{s}{\sqrt{n}}}$

                     t = $\frac{(217.222-216)}{\frac{1.202}{\sqrt{9}}}$

                       = $\frac{1.222}{0.40066}}$

                     t = 3.0509

Critical value = t(a,n-1) = t(0.025,9-1)

                      = 2.306

Since t > critical value, Hence reject H0.

Option b. is the answer. Yes, because computed t is greater than the critical value.