Question

At company XYZ there are 20,000 pilots. We took two random samples, one from the female population, and one from the male population. The following table contains the salary distribution and gender of the pilots.

Salary Female

Salary Female
32,368	53,429	24,967
24,395	58,968	35,423
23,975	27,525	41,188
49,033	44,183	27,525
50,175	24,352	41,188
24,352	47,180	57,242
27,525	53,429	53,429
29,606	26,491	44,138

Salary Male

Salary Male
53,174	24,395	44,884	54,981
52,722	43,124	46,574	62,530
53,423	53,174	53,174	24,395
50,602	58,515	53,627	56,884
49,033	56,294	49,033	52,111
35,018	49,033	53,174	49,033
44,183	46,574	56,294	53,174
35,423	52,722	49,033	53,627
49,033	51,237	55,549	42,961
40,741	53,627	51,237	53,174
37,292	37,292	53,174

Assuming that on average everybody has the same experience and education there should not be any difference in salary between male and female. Let’s test this idea.

The sample mean and standard deviation for female are $ 40,306 and $ 11,658. The sample mean and standard deviation for male are $ 48,727 and $ 8,207. Random sample female has size 31 and random sample male has size 43. We are using a level of significance of .05.

Use the three methods (Confidence Interval Approach, Critical Value Approach, p-Value Approach) to test the Null Hypothesis. Please give a precise explanation - as if you are telling someone who doesn’t know anything about statistics - for the reasoning behind your answer. Show all work. What is your conclusion? What is the implication of your conclusion?

Answer 1

The hypothesis are

H0: μ1=μ2

Ha: μ1≠μ2

Parameters:

α	=	Significance Level
x1	=	Sample Mean 1=40306
x2	=	Sample Mean 2=48727
s1	=	Sample Standard Deviation 1=11658
s2	=	Sample Standard Deviation 2=8207
n1	=	Sample Size 1=31
n2	=	Sample Size 2=43

Rejection Region:
Reject H0 if |Zobs| > Z0.025 = 1.95996

Test Statistic:

Zobs=−3.45211

The p value calculated as P-Value is 0.000556.

P value computed using given table

Conclusion:
Since |Zobs| = 3.45211 > Z0.025 = 1.95996 and the p-value = 0.000556 < α = 0.05 we reject H0 and conclude that at the 5.0 % level of significance there is sufficient evidence to suggest that sample one has a significantly different mean to sample two.

Cinfidence interval approach.

The formula for estimation is:

μ1 - μ2 = (X1 - X2) ± ts(X1 - X2)
where:

X1 & X2 = sample means
t = t statistic determined by confidence level
s(X1 - X2) = standard error = √((s2p/n1) + (s2p/n2))

Pooled Variance
s2p = ((df1)(s21) + (df2)(s22)) / (df1 + df2) = 6906172578 / 72 = 95919063.58

Standard Error
s(X1 - X2) = √((s2p/n1) + (s2p/n2)) = √((95919063.58/31) + (95919063.58/43)) = 2307.56

Confidence Interval
μ1 - μ2 = (X1 - X2) ± ts(X1 - X2) = 8421 ± (1.99 * 2307.56) = 8421 ± 4600.04

Hence confidence interval as

[3820.96, 13021.04].

Since the confidence interval does not include 0 henceforward we can say that we both the means of female nad male are statistically different.