Despite some controversy, scientists generally agree that high-fiber cereals red

Question

Despite some controversy, scientists generally agree that high-fiber cereals reduce the likelihood of various forms of cancer. However, one scientist claims that people who eat high-fiber cereal for breakfast will consume, on average, fewer calories for lunch than people who don’t eat high-fiber cereal for breakfast. If this is true, high-fiber cereal manufacturers will be able to claim another advantage of eating their product – potential weight reduction for dieters. As a preliminary test of the claim, 142 people were randomly selected and asked what they regularly eat for breakfast and lunch. Each person was identified as either a consumer or a non-consumer of high-fiber cereal, and the number of calories consumed at lunch was measured and recorded

The results of the study follow:

Before conducting a test to assess differences in calorie consumption, the analyst wishes to test one of the assumptions. Notably, the equality of variances.

1. Formulate the null and an alternative hypothesis.

2. What are the assumptions, if any, required for this test?

3. Compute the test statistic.

4. Using a level of significance of 10% and the critical value approach for the test, what is your conclusion?

5. What is an approximate p-value for this test?

Consumer Non-consumer 604.2 633.23 Sample Mean Sample Variance Sample Size 4,103 10,670 41 101

Explanation / Answer

here we use t-test with

(1)null hypothesis H0:mean1=mean2 and alternate hypothesis H1:mean1<mean2 ( one tailed)

(2) equlality of variance and sample comes from normal distribution

(3) t=(mean1-mean2)/((sp*(1/n1 +1/n2)1/2) and sp2=((n1-1)s12+(n2-1)s22)/n and with df is n=n1+n2-2

(4) since calcualted t=3.56 is more than one tailed critical t(0.1/2, 140)=1,29 , so we fail to accpt H0 and conclude that consumer mean is less than to non-consumer mean.

(5)

( using ms-excel= =TDIST(3.56,140,1)

t-test sample mean s s2 n (n-1)s2 consumer 604.2 4103 41 164120 non-consumer 633.23 1067 101 106700 difference= 29.03 5170 142 270820 sp2= 1934.428571 sp= 43.98213923 t= 3.564340045 one tailed p-value= 0.000249728

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