Question

QUESTION 2 Use minitab Suppose a sample of 20 students were given a diagnostic test before studying a particular module and then again after completing the module. We want to find out if, in general, our teaching leads to improvements in students' knowledge/skills (i.e. test scores). We can use the results from our sample of students to draw conclusions about the impact of this module in general. 92.MPJ 1) check assumption (fort show your correct 3) find the t-value and p-value: and answer at 2 decimal, for p-value show your correct answer at 3 decimal) 4) conclusion ( show your correct 5) find the confidence of interval for the difference of the mean. answer at 3 decimal)
s.

Answer 1

We have data in paired : we have to use paired sample t test :

and have to used MINITAB :

Stat > Basic Statistics > Paired t test > choose sample 1 and sample 2

here sample 1 = pre mod and sample 2 = post mod

To test :

Ho : μD = 0 vs H1: μD ≠ 0

where μD = mean difference between pre mod and post mod .

it is two tailed test :    

α = 5 % ( assume)  

# 1) Assumption : satisfy ,

## 3 ) find t value and p value  

t value = -3 .23 and p value = 0.004

## 4 ) Decision :

we reject Ho if p value is less than α value using p value approach here p value is less than α value

we reject Ho at given level of significance .

Conclusion :

There is sufficient evidence to conclude that there is population mean difference between two samples at α = 5 % level of significance .

## 5) find the confidence interval for difference of the mean :

(-3.378, -0.722)

Interpretation :

we can say 95 % confident that difference of the mean is lies within  -3.378 to   -0.722 .

interval does not include null value hence we can say it is significant and there is difference in means .

## output of the test from minitab :

Welcome to Minitab, press F1 for help.

MTB > Paired 'Pre mod' 'Post mod';
SUBC> Confidence 95.0;
SUBC> Test 0.0;
SUBC> Alternative 0;
SUBC> GBoxplot.

Paired T-Test and CI: Pre mod, Post mod

Paired T for Pre mod - Post mod

N Mean StDev SE Mean
Pre mod 20 18.400 3.152 0.705
Post mod 20 20.450 4.058 0.907
Difference 20 -2.050 2.837 0.634

95% CI for mean difference: (-3.378, -0.722)

T-Test of mean difference = 0 (vs ≠ 0): T-Value = -3.23 P-Value = 0.004

box plot for difference between pre mod and post mod

: from plot we can say that it is normal distribution , hence here assumption of normal distribution satisfy .

here sample size is less than < 30 hence sample size assumption satisfy .

here all the assumption satisfy .

Boxplot of Differences (with Ho and 95% t-confidence interval for the mean) Ho -7.5 -5.0 -2.5 0.0 2.5 5.0 Differences