Respuesta :
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:
[tex]H_0: \mu = 10941[/tex]
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:
[tex]H_a: \mu > 10941[/tex]
The test statistic is:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, [tex]\sigma[/tex] is the standard deviation and n is the size of the sample.
10941 is tested at the null hypothesis:
This means that [tex]\mu = 10941[/tex]
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 [tex]n = 115, X = 11425, \sigma = 3217[/tex]
Value of the z-statistic:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{11425 - 10941}{\frac{3217}{\sqrt{115}}}[/tex]
[tex]z = 1.61[/tex]
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.
