Question

1. (15) Two samples of girls aged 6 and 7 are with the results shown below. given the Wide Range Achievement Test, Age 6 Age 7 Sample Size 11 15 Mean Score 42 51 Standard Deviation 8 Construct a 95% confidence interval for the difference in mean scores for the two age groups. (10 pts) a. b. Based on the interval from part A, do the test scores differ between the two age groups? Explain your answer. (5 pts)

Answer 1

1.

a.

Formula for Confidence Interval for Difference in two Population means when population Standard deviation are not known and assumed population standard deviations are unequal

2 1-T2)t t(a/2,A) n2 ni + X

|Degrees of Freedom: Asf/n1)+(s3/n2)2 2 n1-1 n2-1

Sample 1 : x₁ :age 6 group

Sample 2 : x₂ :age 7 group

Given
n1 : Sample Size of Sample 1	11
n2 : Sample Size of Sample 2	15
$\overline{x}_{1}$ : Sample Mean of Sample 1	42
$\overline{x}_{2}$: Sample Mean of Sample 2	51
s1 : Sample Standard Deviation of Sample 1	8
s2 : Sample Standard Deviation of Sample 2	6
Confidence Level	95%

|Degrees of Freedom: Asf/n1)+(s3/n2)2 2 n1-1 n2-1

[(82/11)(62/15)] [(5.8182)(2.4)]2 67.5388 17.789817 / 11)2 11-1 (62/15)2 15-1 (82 3.3851 0.4114 3.7965

$\alpha$ for 95% confidence level = (100-95)/100 = 0.05

$\alpha$/2 = 0.05/2=0.025

= 2.1098

ta/2.df= to.025,17

95% confidence interval for the difference in mean score of two age groups

82 62 (1 T2) ta/2, n (42-51) 土2.1098 11 15 n2

=-9 t 2.1098 x 2.8667 9 6.0482 (-15.0482,-2.9518

95% confidence interval for the difference in mean score of two age groups = (-15.0482 , -2.9518)

b.

As the both upper and lower confidence limits are less than zero; There is sufficient evidence to conclude that the test scores do differ between the age groups.

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If the population standard deviations are assumed equal ;

Formula for Confidence Interval for Difference in two Population means

1 1 (x- ta/2(n1+n2-2) x Sp1 n1 n2

(n1-1)s(n2-1)s Pooled Sample Standard deviation sp _ n1n2 2

$s_{p}= \sqrt{\frac{(11-1)8^{2}+(15-1)6^{2}}{11+15-2}}=\sqrt{\frac{1144}{24}}=\sqrt{47.6667}=6.9041$

Degrees of freedom (n1+ n2- 2 11-+15-2 24)

$t_{\alpha /2,df}=t_{0.025,24}=2.0639$

95% confidence interval for the difference in mean score of two age groups

1 1 (x- ta/2(n1+n2-2) x Sp1 n1 n2

(42 51) 2.0639 x 6.9041, -92.0639 x 6.9041 x 0.397 + 11 15

=-9 t 5.6564 (-14.6564, -3.3436)

95% confidence interval for the difference in mean score of two age groups : (-14.6564,-3.3436)