Question

1. To compare the effect of stress in the form of noise on the ability to perform a simple task, 70 subjects were divided into two groups. The first group of 30 subjects acted as a control, while the second group of 40 were the experimental group. Although each subject performed the task in the same control room, each of the experimental group subjects had to perform the task while loud rock music was played. The time to finish the task was recorded for each subject and the following summary was obtained: Control Experimental n 30 40 x 15 minutes 23 minutes s 4 minutes 10 minutes (a) Find a 95% confidence interval for the difference in mean completion times for these two groups. Does the confidence interval suggest that there is a difference in the average time to completion for the two groups? (b) Is there sufficient evidence to indicate that that the average time to complete the task was longer for the experimental “rock music” group? Test at the 1% significance level and state your conclusion.

Answer 1

(a)

n1 = 30

n2 = 40

x1-bar = 15

x2-bar = 23

s1 = 4

s2 = 10

% = 95

Degrees of freedom = n1 + n2 - 2 = 30 + 40 -2 = 68

Pooled s = √(((n1 - 1) * s1^2 + (n2 - 1) * s2^2)/DOF) = √(((30 - 1) * 4^2 + ( 40 - 1) * 10^2)/(30 + 40 -2)) = 8.011021819

SE = Pooled s * √((1/n1) + (1/n2)) = 8.01102181923351 * √((1/30) + (1/40)) = 1.934845588

t- score = 1.995468907

Width of the confidence interval = t * SE = 1.99546890722492 * 1.93484558840761 = 3.860924212

Lower Limit of the confidence interval = (x1-bar - x2-bar) - width = -8 - 3.86092421194868 = -11.86092421

Upper Limit of the confidence interval = (x1-bar - x2-bar) + width = -8 + 3.86092421194868 = -4.139075788

The confidence interval is [-11.86, -4.14]

The above interval does not include 0. This suggests that there is a difference in the average time to completion for the two groups.

(b)

Data:     

n1 = 40    

n2 = 30    

x1-bar = 23    

x2-bar = 15    

s1 = 10    

s2 = 4    

Hypotheses:    

Ho: μ1 ≤ μ2    

Ha: μ1 > μ2    

Decision Rule:    

α = 0.01    

Degrees of freedom = 40 + 30 - 2 = 68

Critical t- score = 2.38244578   

Reject Ho if t > 2.38244578   

Test Statistic:    

Pooled SD, s = √[{(n1 - 1) s1^2 + (n2 - 1) s2^2} / (n1 + n2 - 2)] = √(((40 - 1) * 10^2 + (30 - 1) * 4^2)/(40 + 30 - 2)) = 8.011021819

SE = s * √{(1 /n1) + (1 /n2)} = 8.01102181923351 * √((1/40) + (1/30)) = 1.934845588

t = (x1-bar -x2-bar)/SE = (23 - 15)/1.93484558840761 = 4.134696871

p- value = 0   

Decision (in terms of the hypotheses):

Since 4.13469687 > 2.382445783 we reject Ho and accept Ha

Conclusion (in terms of the problem):

There is sufficient evidence that the average time to complete the task was longer for the experimental “rock music” group.