Hello Chegg, can you please help me with this assignment question. Please provid
ID: 2908825 • Letter: H
Question
Hello Chegg, can you please help me with this assignment question. Please provide a detailed description that explains your logical thought process. Answers without a detailed descriptions are not acceptable.
Follow the prompts. You should include this cover sheet, plus your actual solutions, answers and/or graphs The Statistics Department has an undergraduate program learning objective that focuses on correctly identifying various concepts in designs of experiments. As its benchmark assessment, students take a pre-test and a post- test that bracket the unit of study. Assistant Dean Adele Stacy sampled recent paired scores for nine (9) randomly selected undergraduates. She wondered if scores improved after the routine curricular intervention, Should Dean Adele report a statistically significant finding? Explain. She tested her claims at the 2% level of significance Student Pre-Tes?t A | B | C | D | E | F | G | Hi I 24 2 209 18 25 18 17 Post-Test 2 25 122421 66 18 17 COMMENTS & HINTS: Sounds like a matched pair, wouldn't you say? Use technology to your advantage? Perhaps there an app for this in Excel? Yes, indeed. State the hypotheses; a verbal version is satisfactory; also state the conclusions, both technical and contextual Express the p-value as you findings. Constructing a corresponding confidence interval to affirm vour conclusion can earn bonus credit, though if your testing is one tailed, do create a bogus second tail so to use two-tailed methods.] explain yourExplanation / Answer
Solution:
Here, we have to use matched paired t test for checking the significant difference between population means.
The null and alternative hypotheses for this test are given as below:
Null hypothesis: H0: There is no any significant improvement in the average score of students after the routine curricular intervention.
Alternative hypothesis: Ha: There is a significant improvement in the average score of students after the routine curricular intervention.
H0: µd = 0 vs. Ha: µd > 0
This is a one tailed test. (Upper tailed or right tailed)
We are given
Level of significance = ? = 2% = 0.02
The test statistic formula for this test is given as below:
t = Dbar / [Sd/sqrt(n)]
Calculations for Dbar and Sd are summarised in the following table:
Post test
Pre test
Di
(Di - DBar)^2
25
24
1
0.444444444
21
22
-1
7.111111111
22
20
2
0.111111111
24
19
5
11.11111111
21
18
3
1.777777778
16
12
4
5.444444444
16
15
1
0.444444444
18
18
0
2.777777778
17
17
0
2.777777778
Dbar = 1.666667
Sd = 2.0000
Sample size = n = 9
DF = n – 1 = 9 – 1 = 8
? = 0.02
Upper critical value = 2.4490
(by using t-table or excel)
Test statistic is given as below:
t = 1.666667/(2.0000/sqrt(9))
t = 1.666667/(2/3)
t = 2.500001
Test statistic = t = 2.50
P-value = 0.0185
(by using t-table or excel)
P-value < ? = 0.02
So, we reject the null hypothesis that there is no any significant improvement in the average score of students after the routine curricular intervention.
There is sufficient evidence to conclude that there is a significant improvement in the average score of students after the routine curricular intervention.
Now, we have to construct confidence interval.
We have ? = 0.02, so c = 1 - ? = 1 - 0.02 = 0.98 = 98%
We have to construct 98% confidence interval for difference.
Dbar = 1.666667
Sd = 2.0000
Sample size = n = 9
DF = n – 1 = 9 – 1 = 8
Critical t value = 2.8965
Confidence interval = Dbar ± t*Sd/sqrt(n)
Confidence interval = 1.666667 ± 2.8965*2/sqrt(9)
Confidence interval = 1.666667 ± 2.8965*2/3
Confidence interval = 1.666667 ± 2.8965* 0.666667
Confidence interval = 1.666667 ± 1.931001
Lower limit = 1.666667 - 1.931001 = -0.264334
Upper limit = 1.666667 + 1.931001 = 3.597668
Confidence interval = (-0.264334, 3.597668)
Post test
Pre test
Di
(Di - DBar)^2
25
24
1
0.444444444
21
22
-1
7.111111111
22
20
2
0.111111111
24
19
5
11.11111111
21
18
3
1.777777778
16
12
4
5.444444444
16
15
1
0.444444444
18
18
0
2.777777778
17
17
0
2.777777778
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