Answer
α = 0.01, df = 400-1=399, critical value= 2.326
t=$ \frac{\bar{x}-μ }{σ_\bar{s}}$
= $ \frac{114630-110510 }{66.7509}$
=2.7
Rejection region = z > 2.326
Non rejection region = z < 2.326
The value z = 2.7 falls within the rejection region, hence we reject the null hypothesis and conclude that the current average annual earnings of financial analysts is higher than 110,510 dollars.
b.The probability of making a Type I error : α = 0.01. This is because we can be 99% confident that the test statistics are true, there is a 1% probability of error to accept a null hypothesis that is an error.
c. If the probability of making a Type I error=0,
Rejection region = z > 4
Non rejection region = z < 4
The conclusion of part a will changed if the probability of making a Type I error is zero.
d. By p value approach,
If α = 0.01,
p-value = P(Z>2.7) = 0.0035 < α
The p-value =0.0035 is smaller than α , hence we reject the null hypothesis and conclude that the current average annual earnings of financial analysts is higher than 110,510 dollars.
Work Step by Step
--
