Question

A company claims that the mean monthly residential electricity consumption in a certain region is more than 870 kWh. You want to test this claim. You find that a random sample of 67 residential customers has a mean monthly consumption of 910 kWh. Assume the population standard deviation is 127 kWh. At ?=0.10, can you support the claim? Complete parts (a) through (e).(a) IdentifyUpper H 0H0andUpper H Subscript aHa.(b) Find the critical? value(s) and identify the rejection? region(c) Find the standardized test statistic. Use technology.(d) Decide whether to reject or fail to reject the null hypothesis.

Answer 1

a) Null hypothesis:
`\mu \leq 870`

Alternative hypothesis:
`\mu > 870`

b) The critical region or the rejection zone for the null hypotheiss would be:

`(1.28;\infty)`

c) z=2.578

d) If we compare the p value and the significance level given
`\alpha=0.1` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the monthly residential consumption is more than 870 KWh at 10% of signficance.

Step-by-step explanation:

1) Data given and notation

`\bar X=910` represent the mean of monthly comsumption for the sample

`\sigma=127` represent the population standard deviation for the sample

`n=67` sample size

`\mu_o =870` represent the value that we want to test

`\alpha=0.1` represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

`p_v` represent the p value for the test (variable of interest)

Part a: State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the population mean for the monthly comsumption of electricity is higher than 870, the system of hypothesis would be:

Null hypothesis:
`\mu \leq 870`

Alternative hypothesis:
`\mu > 870`

Since we know the population deviation, and the sample size >30, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

`z=(\bar X-\mu_o)/((\sigma)/(√(n)))` (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Part b: Calculate critical values

Since is a one side upper test we would have just a critical value, and we can calculate from this expression:

`P(z>a)=0.1`

We need a value a such that accumulates 0.1 of the area on the right of the normal standard distribution, and this value is a= 1.28

So the critical region or the rejection zone for the null hypotheiss would be:

`(1.28;\infty)`

Part c: Calculate the statistic

We can replace in formula (1) the info given like this:

`z=(910-870)/((127)/(√(67)))=2.578`

P-value

Since is a one-side upper test the p value would be:

`p_v =P(z>2.578)=0.0049`

In Excel we can use the following formula to find the p value "=1-NORM.DIST(2.578;0;1;TRUE)"

Part d: Conclusion

If we compare the p value and the significance level given
`\alpha=0.1` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the monthly residential consumption is more than 870 KWh at 10% of signficance.