Answer:
The calculated value of t= -1.118860 does not lie in the critical region t= -1.714 or t= -2.5. Therefore we accept our null hypothesis that the mean score for Group 1 is significantly lower than the mean score for Group 2 for both significance level of 0.05 and 0.01
Explanation:
Groups Sample Size Mean Standard Deviation
1 11 65 4.7
2 16 74 2.5
Formulate null and alternate hypotheses as
H0 : u1 < u2 against Ha: u1 ≥ u 2
That is the mean of the group 1 is lower than the mean of the group 2
Degrees of freedom is calculated df = υ= n1+n2- 2= 11+16-2= 23
The significance level alpha is chosen to be ∝ = 0.05
The critical region t ≤ -t (0.05, 23) = -1.714
And if we choose alpha = 0.01 the critical region would be
t ≤ -t (0.05, 23) = -2.5
Here the difference between the sample means is x`1- x`2= 65-74= -9
The pooled estimate for the common variance σ² is
Sp² = 1/n1+n2 -2 [ ∑ (x1i - x1`)² + ∑ (x2j - x`2)²]
= 1/11+16-2 [ (65)²+(74)²]
= 1/23[ 4225+5476]
= 1/23[9701]
=421.782
Sp =√ 421.782=20.5373
The test statistic is
t = (x`1- x` ) /. Sp √1/n1 + 1/n2
t= -9/ 20.5373√1/11+ 1/16
t= -9/8.0439
t= -1.118860
The calculated value of t= -1.118860 does not lie in the critical region t= -1.714 or t= -2.5. Therefore we accept our null hypothesis that the mean score for Group 1 is significantly lower than the mean score for Group 2 for both significance level of 0.05 and 0.01