Answer:
Explanation:
Corresponding test scores of students before and after they completed a formal logic course.
The data for the test are the differences between the test scores.
μd = the score before minus score after
Score before score after diff
74 73 1
83 77 6
75 70 5
88 77 11
84 74 10
63 67 - 4
93 95 - 2
84 83 1
91 84 7
77 75 2
Sample mean, xd
= (1 + 6 + 5 + 11 + 10 - 4 - 2 + 1 + 7 + 2)/10 = 3.7
xd = 3.7
Standard deviation = √(summation(x - mean)²/n
n = 10
Summation(x - mean)² = (1 - 3.7)^2 + (6 - 3.7)^2 + (5 - 3.7)^2+ (11 - 3.7)^2 + (10 - 3.7)^2 + (- 4 - 3.7)^2 + (-2 - 3.7)^2 + (1 - 3.7)^2 + (7 - 3.7)^2 + (2 - 3.7)^2 = 220.1
Standard deviation = √(220.1/10
sd = 4.69
For the null hypothesis
H0: μd ≥ 0
For the alternative hypothesis
H1: μd < 0
The distribution is a students t. Therefore, degree of freedom, df = n - 1 = 10 - 1 = 9
The formula for determining the test statistic is
t = (xd - μd)/(sd/√n)
t = (3.7 - 0)/(4.69/√10)
t = 2.49
The test is a left tailed test. From the t distribution table, the t score corresponding to the significance level is 2.262
In order to reject the null hypothesis, the test statistic must be smaller than - 2.262 or greater than 2.262
Since - 2.49 < - 2.262 and 2.49 > 2.262, we would reject the null hypothesis.
Therefore, at 5% significance level, we can conclude that for the test scores before and after, the mean difference is lesser than zero