Question

A company claims that the mean monthly residential electricity consumption in a certain region is more than

870870

​kiloWatt-hours (kWh). You want to test this claim. You find that a random sample of

7070

residential customers has a mean monthly consumption of

900900

kWh. Assume the population standard deviation is

120120

kWh. At

alphaαequals=0.050.05​,

can you support the​ claim? Complete parts​ (a) through​(e).

​(a) Identify

Upper H 0H0

and

Upper H Subscript aHa.

Choose the correct answer below.

A.

Upper H 0H0​:

muμequals=900900

Upper H Subscript aHa​:

muμnot equals≠900900

​(claim)

B.

Upper H 0H0​:

muμgreater than>900900

​(claim)

Upper H Subscript aHa​:

muμless than or equals≤900900

C.

Upper H 0H0​:

muμgreater than>870870

​(claim)

Upper H Subscript aHa​:

muμless than or equals≤870870

D.

Upper H 0H0​:

muμless than or equals≤870870

Upper H Subscript aHa​:

muμgreater than>870870

​(claim)

E.

Upper H 0H0​:

muμequals=870870

​(claim)

Upper H Subscript aHa​:

muμnot equals≠870870

F.

Upper H 0H0​:

muμless than or equals≤900900

Upper H Subscript aHa​:

muμgreater than>900900

​(claim)

Answer 1

option d) is answer

H0​:μ≤870

Ha​:μ>870 ( )

-------------------

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the critical value for a right-tailed test is z_c = 1.64

The rejection region for this right-tailed test is R={z:z>1.64}

(3) Test Statistics

The z-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that z=2.092>zc​=1.64, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value isp=0.0182, and since p=0.0182<0.05, it is concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ is greater than 870, at the 0.05 significance level.