Question

A consultant for a large university studied the number of hours per week freshmen watch TV versus the number of hours seniors do. The result of this study follow. Is there enough evidence to show the mean number of hours per week freshman watch TV is different from the mean number of hours seniors do at alpha= 0.01?

	Freshmen	Seniors
n	8	4
xbar	18.2	11.9
s	7.8740	3.9749

For the Hypothesis stated above (in terms of Seniors- Freshmen)

What are the critical values?

What is the decision?

What is the p-value? (Round off to 4 decimal place)

Answer 1

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u₁ = u ₂
Alternative hypothesis: u₁ u ₂

Note that these hypotheses constitute a two-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.01. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = sqrt[(s₁²/n₁) + (s₂²/n₂)]
SE = 3.421
DF = 10
t = [ (x₁ - x₂) - d ] / SE

t = 1.84

t_critical = + 3.17

where s₁ is the standard deviation of sample 1, s₂ is the standard deviation of sample 2, n₁ is the size of sample 1, n₂ is the size of sample 2, x₁ is the mean of sample 1, x₂ is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 10 degrees of freedom is more extreme than -1.84; that is, less than -1.84 or greater than 1.84.

Thus, the P-value = 0.096.

Interpret results. Since the P-value (0.096) is greater than the significance level (0.01), we have to accept the null hypothesis.