Question

The average expenditure on Valentine’s Day was expected to be $100.89 (USA Today, February 13, 2006). Do male and female consumers differ in the amounts they spend? The average expenditure in a sample survey of 40 male consumers was $135.67, and the average expenditure in a sample survey of 30 female consumers was $68.64. Based on past surveys, the standard deviation for male consumers is assumed to be $35, and the standard deviation for female consumers is assumed to be $20. Test if there is a difference in spending at 1% level of significance.

Answer 1

Solution

Let X = Expenditure ($) on Valentine’s Day by male consumers

      Y = Expenditure ($) on Valentine’s Day by female consumers

Let mean and standard deviation of X be respectively µ₁ and σ₁ and that of Y be µ₂ and σ₂ .

Given σ₁ = 35 and σ₂ = 20 ....................................................................................................................................... (1)

Hypotheses:

Null: H₀: µ₁ = µ₂ Vs Alternative: H_A: µ₁≠ µ₂

Test Statistic: = 10.110

Z = (Xbar - Ybar)/√{(σ₁²/n₁)+(σ₂²/n₂)} where

Xbar and Ybar are sample averages based on n₁ observations on X and n₂ observations on Y.

Calculations

Summary of Excel calculations is given below:

n1	40
n2	30
Xbar	135.67
Ybar	68.64
σ1	35
σ2	20
σ12	1225
σ22	400
Zcal	10.10994
α	0.01
Zcrit	2.575829
p-value	0
σ12/n1	30.625
σ22/n2	13.33333
Sum	43.95833
Sqrt	6.630108
1- (α/2)	0.995

Distribution, Critical Value and p-value:

Under H₀, Z ~ N(0, 1), where

Hence, for level of significance α%, Critical Value = upper (α/2)% point of N(0, 1) and

p-value = P(Z > | Zcal |).

Using Excel Functions, the above are found to be as shown in the above table:

Decision:

Since | Zcal | > Zcrit, or equivalently, since p-value < α, H₀ is rejected.

Conclusion:

There is sufficient evidence to conclude that there is a difference in spending.  Answer

DONE