A company claims that the mean monthly residential electricity consumption in a certain region is more than 870 kiloWatt-hours (kWh). You want to test this claim. You find that a random sample of 67 residential customers has a mean monthly consumption of 900 kWh. Assume the population standard deviation is 120 kWh. At ? = 0.01, can you support the claim? Complete parts (a) through (e). E. H?: ? > 900 (claim) H?: ? ? 900 F. H?: ? = 900 H?: ? ? 900 (claim) (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. (Round to two decimal places as needed.) A. The critical values are ± [ ]. B. The critical value is 2.33. Identify the rejection region(s). Select the correct choice below. A. The rejection regions are z < - 2.33 and z > 2.33. B. The rejection region is z < 2.33. C. The rejection region is z > 2.33. (c) Find the standardized test statistic. Use technology. The standardized test statistic is z = [ ]. (Round to two decimal places as needed.)
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Step 1: Calculate the standardized test statistic (z-score) using the formula: \[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \] where: - \( \bar{x} = 900 \) (sample mean) - \( \mu = 870 \) (population mean) - \( \sigma = 120 \) (population standard Show more…
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A company claims that the mean monthly residential electricity consumption in a certain region is more than 870 kiloWatt-hours (kWh). You want to test this claim. You find that a random sample of 67 residential customers has a mean monthly consumption of 900 kWh. Assume the population standard deviation is 120 kWh. At α=0.01, can you support the claim? H0: μ≤870 Ha: μ>870 (claim) (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. (Round to two decimal places as needed.) A. The critical values are ± ____ B. The critical value is ______.
Qudsiya A.
A company claims that the mean monthly residential electricity consumption in a certain region is more than 870 kiloWatt-hours (kWh). You want to test this claim. You find that a random sample of 67 residential customers has a mean monthly consumption of 900 kWh. Assume the population standard deviation is 120 kWh. At α = 0.01, can you support the claim? Complete parts (a) through (e). (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. (Round to two decimal places as needed.) The critical value is 2.33. Identify the rejection region(s). Select the correct choice below. A. The rejection regions are z < - 2.33 and z > 2.33. B. The rejection region is z < 2.33. C. The rejection region is z > 2.33.
David N.
The U.S. Energy Information Administration claimed that U.S. residential customers used an average of 10,176 kilowatt hours (kWh) of electricity this year. A local power company believes that residents in their area use more electricity on average than EIA's reported average. To test their claim, the company chooses a random sample of 122 of their customers and calculates that these customers used an average of 10,559kWh of electricity last year. Assuming that the population standard deviation is 2756kWh, is there sufficient evidence to support the power company's claim at the 0.05 level of significance? Step 2 of 3 : Compute the value of the test statistic. Round your answer to two decimal places.
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