Question

Construct a confidence interval for pd, the mean of the differences d for the population of paired data. Assume that the population of paired differences is normally distributed. 5) A test of abstract reasoning is given to a random sample of students before and after they 5) completed a formal logic course. The results are given below. Construct a 95% confidence interval for the mean difference between the before and after scores. Before 74 83 75 88 84 63 93 84 91 77 After 73 77 70 77 74 67 95 83 84 75 A) 1.0<Hd <6.4 B) 0.8 < d < 6.6 C) 1.2 <Hd <5.7 D) 0.2 < d <7.2

Answer 1

Answer:

D) 0.2 <  $\mu_d$ < 7.2

Explanation:

Here the test used is Paired sample t-test.

Confidence interval is given by:

$\overline d \pm t_{(n-1,2 tailed)}*\frac{S_d}{\sqrt n}$

where d_bar is the mean of differences and Sd is standard deviation of differences d.

Sr.No	Before	After	d= Before -After
1	74	73	1
2	83	77	6
3	75	70	5
4	88	77	11
5	84	74	10
6	63	67	-4
7	93	95	-2
8	84	83	1
9	91	84	7
10	77	75	2

Now the mean and standard deviation of d can be calculated manually or by excel.

$\overline d$ = 3.7 and Sd = 4.9452

Also t = 2.262 from table with df= n-1 and two tailed

Now,

$\overline d \pm t_{(n-1,2 tailed)}*\frac{S_d}{\sqrt n} = 3.7 \pm 2.262\frac{4.9452}{\sqrt{10}}=3.7\pm 3.5373$

= ( 3.7 - 3.5373, 3.7+3.5373 ) = ( 0.1626 , 7.2373 )

= ( 0.2 , 7.2 ) ...rounded to 1 decimal.

Hence option D is correct.

D) 0.2 <  $\mu_d$ < 7.2

MINITAB output for reference :

Paired T-Test and CI: Before, After

Paired T for Before - After

N Mean StDev SE Mean
Before 10 81.20 9.07 2.87
After 10 77.50 8.06 2.55
Difference 10 3.70 4.95 1.56

95% CI for mean difference: (0.16, 7.24)
T-Test of mean difference = 0 (vs ≠ 0): T-Value = 2.37 P-Value = 0.042

It gives CI : (0.16, 7.24) = ( 0.2, 7.2 ) rounded to 1 decimal