The NAACP has charged that discrimination exists in the salaries of black versus
ID: 3158920 • Letter: T
Question
The NAACP has charged that discrimination exists in the salaries of black versus white construction workers. They ask you to construct a confidence interval for the mean difference in wages and to choose a confidence level that is favorable to their claim. The (white dominated) union has hired a statistician from Business Statistics to construct a confidence interval which is favorable to their claim that there is no difference in the average wages of black versus white workers. Given the following data: Black: n = 40 sample mean wage = $18,940 S2x = $14,110 White: n = 50 sample mean wage = $19,000 S2x = $13,590 (a) Construct two intervals – one which is a 99% interval and the second which is a 99% interval. (b) What confidence level would be most favorable to the NAACP’s allegations? Why? (c) What confidence level would be most favorable to the union? (d) Now formulate a null and alternative hypothesis and test using the data presented above. EXPLAIN CAREFULLY how you would interpret your results vis-à-vis the NAACP’s allegations.
Explanation / Answer
Solution:
Here, we are given,
n
xbar
S^2
S
40
18940
14110
118.7855
50
19000
13590
116.5762
(a) Construct two intervals – one which is a 95% interval and the second which is a 99% interval.
Solution:
The formula for confidence interval for difference between means is given as below:
Confidence interval = difference -/+ t*standard error
Where difference = x1bar – x2bar
Standard error = sqrt ((S1^2/n1 )+ (S2^2/n2))
d.f. = 50 +40 – 2 = 88
T = 1.9890 for 95% confidence interval
Now, plug all values in the formula
(18940 – 19000) -/+ 1.9890*sqrt((14110/40) + (13590/50))
Lower limit = (18940 – 19000) - 1.9890*sqrt((14110/40) + (13590/50))
Lower limit = -60 - 1.9890*24.99 = -109.7070958
Upper limit = (18940 – 19000) + 1.9890*sqrt((14110/40) + (13590/50))
Upper limit = -60 + 1.9890*24.99 = -10.29290422
Now, for 99% confidence interval, t = 2.6390
Lower limit = (18940 – 19000) – 2.6390*sqrt((14110/40) + (13590/50))
Lower limit = -60 – 2.6390*24.99 = -125.9512447
Upper limit = (18940 – 19000) + 2.6390*sqrt((14110/40) + (13590/50))
Upper limit = -60 + 2.6390*24.99 = 5.951244723
(b) What confidence level would be most favorable to the NAACP’s allegations? Why?
Solution:
The 99% confidence level would be most favourable to the NAACP’s allegations because the width of the 99% confidence interval is more than the width of the 95% confidence interval.
(c) What confidence level would be most favorable to the union?
Solution:
The 95% confidence level would be most favourable to the union because it gives minimum range of difference over the 99% confidence interval.
(d) Now formulate a null and alternative hypothesis and test using the data presented above.
Solution:
Here, we have to use the two sample t test for the difference between two means. The null and alternative hypothesis for this test is given as below:
Null hypothesis: H0: µ1 = µ2
Alternative hypothesis: Ha: µ1 µ2
The test statistic formula is given as below:
Test statistic = t = (x1bar – x2bar) / sqrt ((S1^2/n1 )+ (S2^2/n2))
Test statistic = t = (18940 – 19000) / sqrt((14110/40) + (13590/50))
Test statistic = t = -2.4009
P-value = 0.0186
Here, p-value is less than the given level of significance or alpha value so we reject the null hypothesis there is no significant difference in the two population means.
n
xbar
S^2
S
40
18940
14110
118.7855
50
19000
13590
116.5762