Respuesta :
Answer:
a) 0.0026
P- value is 0.0026
Step-by-step 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
[tex]Z = \frac{x^{-} _{1} -x^{-} _{2} }{\sqrt{\frac{(S.D)_{1} ^{2} }{n_{1} } +\frac{(S.D)_{2} ^{2} }{n_{2} } } }[/tex]
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⁻
[tex]Z = \frac{6.75 -6.25 }{\sqrt{\frac{(1^{2} }{80 } +\frac{((0.95)^{2} }{60} } }[/tex]
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
