A marketing firm asked a random set of married and single me

A marketing firm asked a random set of married and single men as to how much they were willing to spend for a vacation. At = .05, is a difference in the two amounts?

Married men  

Single men  

Sample size  

70  

60  

Mean spending  

380  

385  

Sample variance  

6000  

8000

	Married men	Single men
Sample size	70	60
Mean spending	380	385
Sample variance	6000	8000

Solution

A marketing firm asked a random set of married and single men as to how much they were willing to spend for a vacation.

Let X1 be Marries men.

X2 be single men.

We are given that,

sample size for married men (n1) = 70

sample size for single men (n2) = 60

Mean spending for married men (X1bar) = 380

Mean spending for single men (X2bar) = 385

sample variance for married men (S1²) = 6000

sample variance for married men (S2²) = 8000

sample standard deviation for married men (s1) = sqrt(6000) = 77.4597

sample standard deviation for single men (s2) = sqrt(8000) = 89.4427

Now first we have to check hypothesis regarding variances.

H0 : Variances are equal.

H1 : Variances are not equal.

alpha = 0.05

Test statistic is,

F = larger variance / smaller variance

F = 8000 / 6000 = 1.3333

P-value we can calculate using EXCEL.

syntax :

=FDIST(x, deg_freedom1, deg_freedom2)

where x is test statistic value.

deg_freedom1 = n2 - 1 = 60 - 1 = 59

deg_freedom2 = n1 - 1 = 70 - 1 = 69

P-value = 0.1247

P-value > alpha

Accept H0 at 5% level of significance.

Conclusion : Variances are equal.

So we use pooled variance.

S² = [ (n1-1)*S1² + (n2-1)*S2² ] / (n1+n2-2)

= [(70-1)*6000 + (60-1)*8000] / (70+60-2)

S² = 886000 / 128

S² = 6921.875

S = sqrt(6921.875) = 83.1978

Here the hypothesis for the test is,

H0 : mu1 = mu2 Vs H1 : mu1 mu2

where mu1 and mu2 are two population means.

The test statistic is,

t = (X1bar - X2bar) / S*sqrt[1/n1 + 1/n2]

t = (380 - 385) / 83.1978 * sqrt [1/70 + 1/60 ]

t = -5 / 14.6372

t = -0.3416

P-value syntax :

TDIST(x, deg_freedom, tails)

where x is the absolute value of test statistic.

deg_freedom = n1+n2-2 = 70 + 60 - 2 = 128

tails = 2

P-value = 0.7332

P-value > alpha (0.05)

Accept H0 at 5% level of significance.

Conclusion : The two amouts are same.