Question

A study is being conducted to compare the intake of calories at lunch for consumers of high-fiber breakfast cereals and non-consumers of high fiber breakfast cereals. From a random sample of 43 consumers of high fiber cereals, average calorie intake at lunch was 604.02, with a population standard deviation of 64.05 calories. In a random sample of 107 non-consumers of high fiber breakfast cereals, the average calorie intake at lunch was 633.23, with a population standard deviation of 103.3 calories. Assume that training time for each group is normally distributed. Use the following notations:

μ1: The mean calorie intake at lunch for consumers of high fiber breakfast cereal

μ2: The mean calorie intake at lunch for non-consumers of high fiber breakfast cereal

The goal of the statistical analysis is to determine whether the sample data support the hypothesis that average calorie intake at lunch for consumers of high fiber breakfast cereal is lower than that for non-consumers of high fiber breakfast cereal.

1. What is the p-value?

Select one:

a. 0.02

b. 0.04

c. 0.03

d. 0.98

2. At α=0.1 and using the p-value

Select one:

a. We reject H(0) in favor of H(a)

b. We do not reject H(0)

Answer 1

1) P-value = 0.02

2) We reject H(0) in favor of H(a)

Explanation:

We have to perform a test on the difference between means.

The claim is that average calorie intake at lunch for consumers of high fiber breakfast cereal is lower than that for non-consumers of high fiber breakfast cereal.

Then, the null and alternative hypothesis are:

`H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2 < 0`

being:

μ1: The mean calorie intake at lunch for consumers of high fiber breakfast cereal

μ2: The mean calorie intake at lunch for non-consumers of high fiber breakfast cereal

The sample for the consumers has a size n1=43, a mean of M1=604.02 and a standard deviation of s1=64.05.

The sample for non-consumers has a size n2=107, a mean of M1=633.23 and a standard deviation of s1=103.3.

The difference between means Md is:

`M_d=M_1-M_2=604.02-633.23=-29.21`

The standard error for the difference between means is:

`s_(M_d)=\sqrt{(\sigma_1^2)/(n_1)+(\sigma_2^2)/(n_2)}= \sqrt{(64.05^2)/(43)+(103.3^2)/(107)}= √(95.4047+99.7279)=√(195.1327)\\\\\\s_(M_d)=13.97`

Then, we can calculate the test statistic:

`t=(M_d-(\mu_1-\mu_2))/(s_(M_d)) =(-29.21-(0))/(13.97)=-2.09`

The degrees of freedom are:

`df=n_1+n_2-2=43+107-2=148`

For a t=-2.09 and df=148, the P-value of this left-tailed test is:

`P(t<-2.09)=0.02`

2) If the significance level is 0.1, the P-value is smaller, so the effect is significant. The null hypothesis is rejected.

There is enough evidence to support the claim that average calorie intake at lunch for consumers of high fiber breakfast cereal is lower than that for non-consumers of high fiber breakfast cereal.