Question

1 point) (Hypothetical.) The average salary of all residents of a city is thought to be about $39,000. A research team surveys a random sample of 200 residents; it happens that the average salary of these 200 is about $40,000 with a SD of $12,000. Make a z-test of the null hypothesis that this difference was just chance (in the sampling).

Answer 1

We accept H₀

We don´t have enough evidence to reject H₀

Explanation:

Nomal Distribution

population mean μ₀ = 39000

sample size n = 200

sample mean μ = 40000

sample standard deviation s = 12000

Test hypothesis

As we are just interested in look if the difference was just chance, we will do a one tail-test (right)

1.- Hypothesis

H₀ null hypothesis μ₀ = 39000

Hₐ Alternative hypothesis μ₀ > 39000

2.-We considered the confidence interval of 95 % then

α = 0,05 and z(c) = 1.64

3.-Compute z(s)

z(s) = [ ( μ - μ₀ ) ] / 12000/√200 z(s) = [ 40000- 39000)* √200] / 12000

z(s) = 1000*14,14/ 12000 z(s) = 1.1783

4.-Compare

z(s) and z(c)

1.1783 < 1.64

z(s) is inside acceptance region we accep H₀