16. A company claims that the mean monthly residential electricity consumption i

Question

16. A company claims that the mean monthly residential electricity consumption in a certain region is more than 890 (kWh). You want to test this claim. You find that a random sample of 65 residential customers has a mean monthly consumption of 910 kWh. Assume the population standard deviation is 130 kWh. At =0.05, 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.

(b) Find the critical value(s) and identify the rejection region(s). Select the correct choice below and fill in the answer box within your choice. Use technology.

(c) Identify the rejection region(s). Select the correct choice below.

(d) Decide whether to reject or fail to reject the null hypothesis.

(e) Interpret the decision in the context of the original claim.

17. Find the critical value(s) and rejection region(s) for the indicated t-test, level of significance

, and sample size n. Right-tailed test, =0.005, n=15

The critical value(s) is/are

(Round to the nearest thousandth as needed. Use a comma to separate answers as needed.)

18. Find the critical value(s) and rejection region(s) for the indicated t-test, level of significance , and sample size n.

Two-tailed test, =0.10, n=24

The critical value(s) is/are

(Round to the nearest thousandth as needed. Use a comma to separate answers as needed.)

18. Use a t-test to test the claim about the population mean at the given level of significance using the given sample statistics. Assume the population is normally distributed.

Claim: =51, 500; =0.10

Sample statistics: x=50, 479, s=2500, n=19

Explanation / Answer

Q. 16 (a)

H0 : mean monthly residential electricity consumption <= 890 Kwh. < = 890 KWh

Ha : mean monthly residential electricity consumption > 890 Kwh . > 890 KWh

(b) Here Test Statistic

Here population standard deviation is given so we will use Z test here.

Z = (xbar - )/ (s/n) = (890 - 910)/ (130/ 65)

Z = -20/ 16.1245 = - 1.24

Here critical value of Z = 1.645

(c) so Rejection region Z >=c would be c = 890 + 1.645 * (130/ 65) = 916.525

(d) so the given consumption is under the rejection region so we can no reject the null hypothesis.

(f) And we can conclude that mean monthly residential electricity consumption of that certain region is not more than 890 Kwh.

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