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 greater than the critical value.
B)Yes,because computed t is less 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.

Question

Quizplus · Accepted Answer

We need to perform a one-sample t-test to determine if the mean shelf life of the revised cake mix is greater than the original mean shelf life of 216 days.

The null hypothesis is:
H0: µ = 216
The alternative hypothesis is:
Ha: µ > 216

Since we don't know the population standard deviation, we use a t-test. We calculate the sample mean and standard deviation as follows:

sample mean = (215+217+218+219+216+217+217+218+218)/9 = 217
sample standard deviation = 1.069

The t-score is calculated as:

t = (sample mean - hypothesized mean) / (sample standard deviation / sqrt(sample size))
t = (217 - 216) / (1.069 / sqrt(9))
t = 2.12

The degrees of freedom for this test is n-1=8, and the critical value for a one-tailed test at α = 0.025 with 8 degrees of freedom is 2.306. Since our calculated t-score (2.12) is less than the critical value (2.306), we cannot reject the null hypothesis.

Therefore, there is not enough evidence to conclude that the shelf life of the cake mix has increased at α = 0.025.

A Manufacturer Wants to Increase the Shelf Life of a Line