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r-ruslan [8.4K]
3 years ago
15

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
Mathematics
1 answer:
Umnica [9.8K]3 years ago
4 0

Answer:

a) 0.0026

P- value is <em>0.0026</em>

Step-by-step explanation:

<u><em>Step(i):</em></u>-

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

<u><em>step(ii):-</em></u>

<em>Test statistic </em>

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

<em>   </em><u><em>Null Hypothesis :H₀</em></u><em>: There is no significant difference in wages across the two employers.</em>

<em> x⁻₁= x₂⁻</em>

<em>Alternative Hypothesis :H₁: There is significant difference in wages across the two employers.</em>

<em>x⁻₁≠ x₂⁻</em>

<em></em>

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

<em>Z = 3.01</em>

<u><em>P- value:-</em></u>

Given data is two tailed test

The test statistic Z = 3.01

First we have to find the Probability of z-statistic

<em>P(Z>3.01) =  1- P( z <3.01)</em>

<em>                  = 1- (0.5 + A(3.01)</em>

<em>                 = 0.5 - A(3.01)</em>

<em>              =    0.5 - 0.49865   ( from normal table)</em>

<em>              = 0.0013</em>

<em>P(Z>3.125) = 0.0013</em>

<em>Given two tailed test </em>

<em>    P- value = 2 × P( Z > 3.01)</em>

<em>                  = 2 × 0.0013</em>

<em>                 = 0.0026</em>

<u><em>Final answer</em></u><em>:-</em>

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

<em>null hypothesis is rejected</em>

<u><em>Conclusion:-</em></u>

P- value is <em>0.0026</em>

               

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