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

Answer:

We accept  H₀

We don´t have enough evidence to reject H₀

Step-by-step 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