Answer
$\bar{x}$ = 1198 dollars μ=1170 dollars,σ=125 dollars,$σ_\bar{s}$$ \frac{125}{$100}$=12.5, n = 100
α = 0.01/2=0.005, df = 100-1=99, critical value=2.576
t=$ \frac{\bar{x}-μ }{σ_\bar{s}}$
= $ \frac{1198-1170 }{12.5}$
=2.24
Rejection region = z > 2.576 or z 1.96
Non rejection region = z < 1.96
The value z = 2.24 falls within the rejection region, hence we reject the null hypothesis and claim that the current average such premium paid by all such workers in this city is higher than 1170 dollars.
The probability of making a Type I error : α
Part a : α=1%, Part b: α = 2.5%
If α = 0.01,
p-value = P(Z > 2.24) = 0.0125 > α
The p-value =0.0125 is greater than α ,hence we failed to reject the null hypothesis and claim that the current average such premium paid by all such workers in this city is no different from 1170 dollars.
If α = 0.025,
p-value = P(Z>2.24) = 0.0125 < α
The p-value =0.0125 is smaller than α , hence we reject the null hypothesis and claim that the current average such premium paid by all such workers in this city is higher than 1170 dollars.
Work Step by Step
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