Answer:
The p-value is;
d. 0.00131
Step-by-step explanation:
The given question parameters are;
Company A [tex]{}[/tex] Company B
Sample size 80[tex]{}[/tex] 60
Sample mean [tex]{}[/tex] [tex]\overline x_1[/tex] $16.75 [tex]\overline x_2[/tex] $16.25
Population standard deviation [tex]{}[/tex] σ₁ $1.00 σ₂ $0.95
[tex]z=\dfrac{(\bar{x}_{1}-\bar{x}_{2})-(\mu_{1}-\mu _{2} )}{\sqrt{\dfrac{\sigma_{1}^{2} }{n_{1}}+\dfrac{\sigma _{2}^{2}}{n_{2}}}}[/tex]
Therefor, we have;
[tex]z=\dfrac{(16.75-16.25)-(0-0)}{\sqrt{\dfrac{1^{2} }{80}+\dfrac{0.95^{2}}{60}}} \approx 3.0128[/tex]
The p-value = 1 - 0.99869 (the area to the right of 3.01) = 0.00131
