Qmet 252

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. At the 0.025 level, has the shelf life of the cake mix increased?
Choose one answer.
|[pic]|a. No, because computed t lies in the region of acceptance. [pic] | |
|[pic]|b. Yes, because computed t is less than the critical value. [pic] | |
|[pic]|c. Yes, because computed t is greater than the critical value. [pic] | |
|[pic]|d. No, because 217.24 is quite close to 216. [pic] | |

Population standard deviation unknown, we need to use t test. Using data analysis, we can easily get the sample mean 217.222 and sample standard deviation 1.202.

Ho: u 216
Reject Ho if t > 2.306 (one-tailed test with a = 0.025 and df = 8) t = (217.222 – 216) / (1.202/sqrt(9)) = 3.05
Reject Ho. The shelf life of the cake mix has increased.

Correct
Marks for this submission: 2/2.
Question 2
Marks: 1
If α = 0.05 for a two-tailed test, how large is the acceptance area?
Choose one answer.
|[pic]|a. 0.025 [pic] | |
|[pic]|b. 0.975 [pic] | |
|[pic]|c. 0.05 [pic] | |
|[pic]|d. 0.95 [pic] | |

Incorrect
Marks for this submission: 0/1.
Question 3
Marks: 1
A random sample of size 15 is selected from a normal population. The population standard deviation is unknown. Assume that a two-tailed test at the 0.10 significance level is to be used. For what value of t will the null hypothesis not be rejected?
Choose one answer.
|[pic]|a. To the left of -1.345 or to the right of 1.345 [pic] | |
|[pic]|b. To the left of -1.645 or to the right of 1.645 [pic] | |