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:  a) Null hypothesis:[tex]\mu \leq 870[/tex]  Alternative hypothesis:[tex]\mu > 870[/tex]  b) The critical region or the rejection zone for the null hypotheiss would be:[tex] (1.28;\infty)[/tex]c) z=2.578d) If we compare the p value and the significance level given [tex]\alpha=0.1[/tex] we see that [tex]p_v<\alpha[/tex] 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  [tex]\bar X=910[/tex] represent the mean of monthly comsumption for the sample  [tex]\sigma=127[/tex] represent the population standard deviation for the sample  [tex]n=67[/tex] sample size  [tex]\mu_o =870[/tex] represent the value that we want to test  [tex]\alpha=0.1[/tex] represent the significance level for the hypothesis test.  z would represent the statistic (variable of interest)  [tex]p_v[/tex] 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:[tex]\mu \leq 870[/tex]  Alternative hypothesis:[tex]\mu > 870[/tex]  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:  [tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] (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 valuesSince is a one side upper test we would have just a critical value, and we can calculate from this expression:[tex]P(z>a)=0.1[/tex]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:[tex] (1.28;\infty)[/tex]Part c: Calculate the statistic  We can replace in formula (1) the info given like this:  [tex]z=\frac{910-870}{\frac{127}{\sqrt{67}}}=2.578[/tex]  P-value  Since is a one-side upper test the p value would be:  [tex]p_v =P(z>2.578)=0.0049[/tex]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 [tex]\alpha=0.1[/tex] we see that [tex]p_v<\alpha[/tex] 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.