Question

The American Society for Quality (ASQ) conducted a salary survey of all its members. ASQ members work in all areas of manufacturing and service-related institutions, with a common theme of an interest in quality. Two job titles are master black belt and green belt. (These are titles established by a Six Sigma quality improvement initiative.) Descriptive statistics concerning salaries for these two job titles are given in the following table:

job title	sample size	mean	standard Deviation
black belt	141	88,945	21,791
green belt	26	64,794	25,911

a. Using a 0.05 level of significance, is there a difference in the variability of salaries between black belts and green belts?

b. Based, on the results of (a), which T-Test would you use to appropriately compare mean salaries?

c. Using, a 0.05 level of significance, is the mean salary of black belts greater than the mean salary of green belts?

Answer 1

a.

We need to check for homogeneity of variances.

H0:

H1:

Test statistic, F = s2² / s1²   = 25911² / 21791²  = 1.414

Numerator df = n2 - 1 = 26 - 1 = 25

Denominator df = n1 - 1 = 141 - 1 = 140

Critical value of F at 0.05 significance level and df = 25, 140 is 1.58

Since the observed F (1.414) is less than the critical value, we conclude that there is no significant evidence of a difference in the variability of salaries between black belts and green belts.

b.

Since, we assume that the population standard deviations of both groups to be equal and we can apply Pooled t-test.

c.

Let and be the mean salary of black belts and green belts respectively.

H0:

H1:

Pooled variance Sp² = [(n1 - 1) s1²   + (n2 - 1)s2² ] / (n1 + n2 - 2)

= [(141 - 1) 21791² + (26 - 1) 25911² ] / (141 + 26 -2)

= 504625293

Standard error of mean difference = sqrt( Sp² * ((1/n1) + (1/n2)))

= sqrt( 504625293 * ((1/141) + (1/26)))

= 4794.535

Test statistic, t = (x1 - x2) / Std error

= (88945 - 64794) / 4794.535

= 5.037

Degree of freedom = n1 + n2 - 2 = 141 + 26 - 2 = 165

P-value = P(t > 5.037, df = 165) = 0.0000

Since p-value is less than 0.05 significance level, we reject null hypothesis H0 and conclude that there is significant evidence that the mean salary of black belts greater than the mean salary of green belts.