Question

Two random samples were drawn from two employers to obtain information about hourly wages. Use the following information and the PopMeanDiff template to determine if there is a significant difference in wages across the two employers.

Kroger Wal-Mart
Sample size 80 60
Sample mean $6.75 $6.25
Population standard deviation $1.00 $0.95

The p-value is _____.

a. 0.0026
b. 0.0013
c. 0.0084
d. 0.0042

Answer 1

a) 0.0026

P- value is 0.0026

Explanation:

Step(i):-

Given data

first sample size n₁= 80

mean of the first sample x⁻₁= $6.75

Standard deviation of the first sample (σ₁) = $1.00

second sample size (n₂) = 60

mean of the second sample( x₂⁻) = $6.25

Standard deviation of the second sample (σ₂) = $0.95

step(ii):-

Test statistic

`Z = \frac{x^(-) _(1) -x^(-) _(2) }{\sqrt{((S.D)_(1) ^(2) )/(n_(1) ) +((S.D)_(2) ^(2) )/(n_(2) ) } }`

Null Hypothesis :H₀: There is no significant difference in wages across the two employers.

x⁻₁= x₂⁻

Alternative Hypothesis :H₁: There is significant difference in wages across the two employers.

x⁻₁≠ x₂⁻

`Z = \frac{6.75 -6.25 }{\sqrt{((1^(2) )/(80 ) +(((0.95)^(2) )/(60) } }`

Z = 3.01

P- value:-

Given data is two tailed test

The test statistic Z = 3.01

First we have to find the Probability of z-statistic

P(Z>3.01) = 1- P( z <3.01)

= 1- (0.5 + A(3.01)

= 0.5 - A(3.01)

= 0.5 - 0.49865 ( from normal table)

= 0.0013

P(Z>3.125) = 0.0013

Given two tailed test

P- value = 2 × P( Z > 3.01)

= 2 × 0.0013

= 0.0026

Final answer:-

The calculated value Z = 3.125 > 1.96 at 0.05 level of significance

null hypothesis is rejected

Conclusion:-

P- value is 0.0026