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?

• - No, because computed t lies in the region of acceptance.

• - No, because 217.24 is quite close to 216.

• - Yes, because computed t is greater than the critical value.

• - Yes, because computed t is less than the critical value.

Answer 1

Answer:

Given,

Mean = $\sum$ x/n

= (215 + 217 + 218 + 219 + 216 + 217 + 217 + 218 + 218)/9

= 217.22

Standard deviation = sqrt($\sum$(x-xbar)^2/(n-1))

substitute values

= 1.202

consider,

test statistic t = (x-u)/(s/sqrt(n))

substitute values

= (217.22 - 216)/(1.202/sqrt(9))

t = 3.045

degree of freedom = n - 1

= 9 - 1

= 8

P value = 0.007971

α = 0.025, df = 8

tcritical value = 2.306

Here we observe that,

test statistics is higher than the critical value , so we reject Ho.

Hence we have sufficient evidence.

So the shelf has increased.