Question

When only two treatments are involved, ANOVA and the Student’s t test (Chapter 11) result in the same conclusions. Also, for computed test statistics, t2 = F. To demonstrate this relationship, use the following example. Fourteen randomly selected students enrolled in a history course were divided into two groups, one consisting of 6 students who took the course in the normal lecture format. The other group of 8 students took the course as a distance course format. At the end of the course, each group was examined with a 50-item test. The following is a list of the number correct for each of the two groups.

Traditional lecture: 36, 31, 40, 34, 33, 38

Distance: 45, 35, 45, 35, 45, 36, 43, 41

Complete the ANOVA table. (Round your SS, MS, and F values to 2 decimal places and p value to 4 decimal places.)

	Source SS df MS F p Factors 1 Error 12 Total 13

Use a α = 0.01 level of significance - the critical value of F:

Using the t test from Chapter 11, compute t.

Answer 1

Solution:-

a)

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

Null hypothesis: u₁ = u₂

Alternative hypothesis: At-least one of the u is not equal.

Formulate an analysis plan. For this analysis, the significance level is 0.01.

Analyze sample data.

F statistics is given by:-

$F=\frac{MSB}{MSE}$

F = 5.67

F_critical = 9.298

The P-value = 0.035

Interpret results. Since the P-value (0.35) is greater than the significance level (0.01), we have to accept the null hypothesis.

Conclusion:-

Reject H₀, There is sufficient evidence for significant differences between the three kinds.

b)

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

Null hypothesis: u₁ = u ₂
Alternative hypothesis: u₁$\neq$ u 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 = 2.1178
DF = 12
t = [ (x₁ - x₂) - d ] / SE

t = -2.499

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 12 degrees of freedom is more extreme than -2.499; that is, less than -2.499 or greater than 2.499.

Thus, the P-value = 0.028.

Interpret results. Since the P-value (0.028) is greater than the significance level (0.01), we have to accept the null hypothesis.

From the above test we do not have sufficient evidence in the favor of the claim that u₁ = u ₂ .