Test the given claim about the means of two populations. Assume that two dependent samples have been randomly selected from normally distributed populations. A test of abstract reasoning is given to a random sample of students before and after they completed a formal logic course. The results are given below. At the 0.05 significance level, test the claim that the mean score is not affected by the course. Assume all requirements are met. Use the critical value method of testing hypotheses.
Before 74 83 75 88 84 63 93 84 91 77
After 73 77 70 77 74 67 95 83 84 75
A. State the null hypothesis (H_0) and the alternate (H_1).
B. What is the test statistic and the critical value(s.
C. Find the confidence interval values.
D. Interpret the information and state what can you say about the claim?

Step-by-step 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

no pictures of the graph

both the histogram and the stem and leaf plot will visually show how many shoppers were of various ages within each range of data.   the histogram does not give us the specific ages of these shoppers unlike the stem and leaf plot.a histogram does not give us a true range of the data because we do not know exactly how old the shoppers were. a stem and leaf plot will give exact ages, so we can determine the exact range along with all measures of center and the others measures of center.   with the histogram, we cannot find any of these.

step-by-step explanation: