Question

10. To test whether a college course is working a pre- and post-test is arranged for the students. The results are given below. Compare the scores with a t-test as indicated. Use α 0.05 to test the claim. Use the 5-step method. a) Assuming the scores are randomly selected from the two groups. Assume equal variances. b) Assume that the scores are pairs of scores for ten students Test Pre-test Post-test 56 8874 64 83 Scores 60 68 57 58 53 72 50 62 64 74 66v 60 65 67

I have the answers but I need to know the steps.

Answer 1

a)

Before After 679 65.2 67.9 7.8711851430 12.05035730 652 Sum Mean S.D Count S.E 10 10 2.4890872936 3.810657569

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. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. 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 = 4.552
DF = 18
t = [ (x₁ - x₂) - d ] / SE

t = -0.593

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 18  degrees of freedom is more extreme than -0.593; that is, less than -0.593 or greater than 0.593.

Thus, the P-value = 0.568

Interpret results. Since the P-value (0.568) is greater than the significance level (0.05), we have to accept the null hypothesis.

From the above test we do not have sufficient evidence in the favor of the claim that there is significance difference.

b)

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

Null hypothesis: u_d = 0

Alternative hypothesis: u_d ≠ 0

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a matched-pairs t-test of the null hypothesis.

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

$s = \sqrt{\frac{\sum (d-\bar{d})^{2}}{n-1}}$

s = 8.52513

SE = s / sqrt(n)

S.E = 2.6959

DF = n - 1 = 10 -1

D.F = 9

t = [ (x₁ - x₂) - D ] / SE

t = -1.001

where d_i is the observed difference for pair i, d is mean difference between sample pairs, D is the hypothesized mean difference between population pairs, and n is the number of pairs.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 9 degrees of freedom is more extreme than -1.001; that is, less than - 1.001 or greater than 1.001.

Thus, the P-value = 0.343

Interpret results. Since the P-value (0.343) is greater than the significance level (0.05), we have to accept the null hypothesis.

Do not reject H0. The mean difference appears does not differ from zero.