THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION:
A restaurant manager was in charge of purchasing new microwaves for his restaurant.The choices were narrowed to two available models.Since the two models cost about the same,the manager was interested in determining whether there was a difference in the amount of power consumed when operated at high voltages.Based on two independent random samples,the following summary information was computed:

-Find a 95% confidence interval for the difference in the amount of power consumed at high voltages by the two models.
A)0.2 ± 0.49
B)0.2 ± 0.58
C)0.2 ± 0.64
D)0.2 ± 1.28

Question

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION:
A restaurant manager was in charge of purchasing new microwaves for his restaurant.The choices were narrowed to two available models.Since the two models cost about the same,the manager was interested in determining whether there was a difference in the amount of power consumed when operated at high voltages.Based on two independent random samples,the following summary information was computed:
 
-Find a 95% confidence interval for the difference in the amount of power consumed at high voltages by the two models.
A)0.2 ± 0.49 
B)0.2 ± 0.58 
C)0.2 ± 0.64 
D)0.2 ± 1.28

Quizplus · Accepted Answer

The 95% confidence interval for the difference in the amount of power consumed at high voltages by the two models is given by: 
(0.2 ± t_s*SE)
where t_s is the t-value with (n1 + n2 - 2) degrees of freedom and a level of significance of 0.05/2 = 0.025, and SE is the standard error of the difference in the sample means. 
From the information given, we have: 
n1 = n2 = 20 (sample sizes for both models)
x̄1 = 1250 W, s1 = 50 W (sample mean and standard deviation for model 1)
x̄2 = 1300 W, s2 = 40 W (sample mean and standard deviation for model 2)
SE = sqrt(s1^2/n1 + s2^2/n2) = sqrt(50^2/20 + 40^2/20) = 11.18 W
The t-value for a 95% confidence interval with 38 degrees of freedom is 2.024 (obtained from a t-table or calculator)
Thus, the 95% confidence interval is: 
0.2 ± 2.024*11.18 = (0.2 ± 22.63)
Which simplifies to (approximately): 
0.2 ± 0.58 
Therefore, the best choice is B.