Question

17. Nutraloaf is a food served in prison to inmates who have misbehaved. The ingredients of nutraloaf vary on a state- by-state basis; consequently, the nutrition facts vary as well. Nationwide, the average number of calories in one serving of nutraloaf is assumed to be 1000. However, a random sample of 20 servings of Vermont nutraloaf yielded an average of 966 calories. Suppose we want to determine if there is enough evidence to conclude that the average number of calories in a serving of Vermont nutraloaf is different than 1000. In a randomization distribution for a test of ?0: ? = 1000 vs. ??: ? ≠ 1000, the following statements are found to be true: (1) the proportion of randomization sample means less than or equal to 966 is .007, and (2) the proportion of randomization sample means greater than or equal to 1000 is .489. Based on the randomization distribution, what is a correct p-value for the hypothesis test?

    a. .014 ,b. .978 ,c. .489 ,d. .007

Answer 1

Solution:

We are given that: Nationwide, the average number of calories in one serving of nutraloaf is assumed to be 1000.

That is:

A random sample of 20 servings of Vermont nutraloaf yielded an average of 966 calories.

Sample size = n = 20

Sample mean =

We want to determine if there is enough evidence to conclude that the average number of calories in a serving of Vermont nutraloaf is different than 1000.

In a randomization distribution for a test of ?0: ? = 1000 vs. ??: ? ≠ 1000,

the following statements are found to be true:

(1) the proportion of randomization sample means less than or equal to 966 is 0.007, and

(2) the proportion of randomization sample means greater than or equal to 1000 is 0.489

For two tailed test ,

a) P-value is 2 times proportion of randomization sample means less than or equal to Sample mean value, if sample mean is less than population mean and

b) P-value is 2 times proportion of randomization sample means greater than or equal to Sample mean value, if sample mean is greater than population mean

Since sample mean is 966 < 1000, so we use part a) to get P-value.

We know the proportion of randomization sample means less than or equal to 966 is 0.007,

Thus P-value = 2 times 0.007 = 0.014

Thus P-value = 0.014 is correct.