The NAACP has charged that discrimination exists in the salaries of black versus

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

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