Question

The U.S. Energy Information Administration claimed that U.S. residential customers used an average of 10,941 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 115 of their customers and calculates that these customers used an average of 11,425kWh of electricity last year. Assuming that the population standard deviation is 3217kWh, is there sufficient evidence to support the power company's claim at the 0.02 level of significance

Answer 1

Answer:

The pvalue of the test is 0.0537 > 0.02, which means that there is not sufficient evidence to support the power company's claim at the 0.02 level of significance.

Step-by-step explanation:

The U.S. Energy Information Administration claimed that U.S. residential customers used an average of 10,941 kilowatt hours (kWh) of electricity this year.

This means that the null hypothesis is:

$H_0: \mu = 10941$

A local power company believes that residents in their area use more electricity on average than EIA's reported average.

This means that the alternate hypothesis is:

$H_a: \mu > 10941$

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation and n is the size of the sample.

10941 is tested at the null hypothesis:

This means that $\mu = 10941$

To test their claim, the company chooses a random sample of 115 of their customers and calculates that these customers used an average of 11,425kWh of electricity last year. The population standard deviation is of 3217kWh:

This means that $n = 115, X = 11425, \sigma = 3217$

Value of the z-statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{11425 - 10941}{\frac{3217}{\sqrt{115}}}$

$z = 1.61$

Pvalue of the test:

Probability of finding a mean above 11425, which is 1 subtracted by the pvalue of z = 1.61.

Looking at the z-table, z = 1.61 has a pvalue of 0.9463.

1 - 0.9463 = 0.0537

The pvalue of the test is 0.0537 > 0.02, which means that there is not sufficient evidence to support the power company's claim at the 0.02 level of significance.