Question

Another ttest question with the pre and post test copmpasite scores

pre

49
67
61
61
64
58
58
61
49
67
55
55
58
67
52
43
58
58
70
52
64
61
58

Post

55
70
64
67
73
88
67
82
82
73
76
88
55
88
58
76
76
76
66
76
76
61
82

Answer 1

Solution:-

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).

	Before	After	d = A - B	(d - dbar)^2
	49	55	-6	68.96219282
	67	70	-3	127.7882798
	61	64	-3	127.7882798
	61	67	-6	68.96219282
	64	73	-9	28.13610586
	58	88	-30	246.3534972
	58	67	-9	28.13610586
	61	82	-21	44.83175803
	49	82	-33	349.5274102
	67	73	-6	68.96219282
	55	76	-21	44.83175803
	55	88	-33	349.5274102
	58	55	3	299.4404537
	67	88	-21	44.83175803
	52	58	-6	68.96219282
	43	76	-33	349.5274102
	58	76	-18	13.65784499
	58	76	-18	13.65784499
	70	66	4	335.0491493
	52	76	-24	94.00567108
	64	76	-12	5.310018904
	61	61	0	204.6143667
	58	82	-24	94.00567108
Sum	1346	1675	-329	3076.869565
Mean	58.521739	72.82608696	-14.304348	133.7769376

s = sqrt [ (\sum (d_i - d)² / (n - 1) ]

s = 11.826

SE = s / sqrt(n)

S.E = 2.4659

DF = n - 1 = 23 -1

D.F = 22

t = [ (x₁ - x₂) - D ] / SE

t = - 5.80

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 22 degrees of freedom is more extreme than - 5.80; that is, less than - 5.80 or greater than 5.80.

Thus, the P-value = less than 0.0001

Interpret results. Since the P-value (0.00) is less than the significance level (0.05), we have to reject the null hypothesis.

Reject H0. The mean difference appears to differ from zero.