please help me solve and show me how to solve A consultant for a large universit

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please help me solve and show me how to solve

A consultant for a large university studied the number of hours per week freshmen watch TV versus the number of hours seniors do. The results of this study follow. Is there enough evidence to show the mean number of hours per week freshmen watch TV is different from the mean number of hours seniors do at -0.1? 61 20 FreshmenSeniors 21 xbar 18.3 11.4 23 7.8740 3.9749 24 25 26 For the hypothesis stated above (in terms of Seniors- Freshmen): 27 28 29 30 1 Question 1What is/are the critical value(s)? Question 2 What is the decision? What is the p-value? Fill in only one of the following statements. If the 2 table is appropriate Question a p-value

Explanation / Answer

Solution:-

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

Null hypothesis: 1 = 2

Alternative hypothesis: 1 2

Note that these hypotheses constitute a two-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.10. 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 = 3.292

DF = 11

t = [ (x1 - x2) - d ] / SE

t = 2.096

tcritical = + 1.796

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 11 degrees of freedom is more extreme than -2.096; that is, less than -2.096 or greater than 2.096.

Thus, the P-value = 0.06

Interpret results. Since the P-value (0.06) is less than the significance level (0.10), we cannot accept the null hypothesis.

From the above test we have sufficient evidence that mean number of hours per week freshman watch TV is different from the mean number of hours senior do at alpha = 0.10.

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