2. I want to compare STAT 200 students’ scores on the Lesson 1 Quiz and the Less

Question

2. I want to compare STAT 200 students’ scores on the Lesson 1 Quiz and the Lesson 2 Quiz. In a random sample of 40 students, the mean difference was 0.200 points with a standard deviation of 2.089 points. Scored tended to be higher on the Lesson 1 Quiz. Use Minitab Express to answer the following questions. You will have to enter in these statistics as “summarized data.”

A. Construct a 95% confidence interval for the mean difference between the scores on the two quizzes.

B. Use the five-step hypothesis testing procedure to test the alternative hypothesis that scores are different on the two quizzes.

Step 1:

Step 2:

Step 3:

Step 4:

Step 5:

Explanation / Answer

a) Margin of error = (2.023*2.089)/Sqrt(40) = 0.6681

95% Interval

(0.2-0.6681, 0.2+0.6681)

= (-0.4681, 0.8681)

So, 95% Interval is (-0.4681, 0.8681)

b) Hypothesis

Null hypothesis : H0: ud = 0

Alternate Hypothesis: H1: ud is not equal to zero

Test Statistic = (0.2/2.089)*Sqrt(40) = 0.6056

P value = P(t<-0.6056)+P(t>0.6056) = 0.5483

Since P value is greater than 0.05, we failed to reject the null hypothesis and conclude that there is mean different scores are not different

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