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abruzzese [7]
2 years ago
7

1. The U.S. Energy Information Administration claimed that U.S. residential customers used an average of 10,290 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 153 of their customers and calculates that these customers used an average of 10,586kWh of electricity last year. Assuming that the population standard deviation is 2509kWh, is there sufficient evidence to support the power company's claim at the 0.05 level of significance? State the null and alternative hypotheses for the test.
2. The U.S. Energy Information Administration claimed that U.S. residential customers used an average of 10,290 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 153 of their customers and calculates that these customers used an average of 10,446 kWh of electricity last year, Assuming that the population standard deviation is 2509 kWh, is there sufficient evidence to support the power company's claim at the 0.02 level of significance? Compute the value of the test statistic.
A. We reject the null hypothesis and conclude that there is insufficient evidence at a 0.02 level of significance to support the power company's claim that the mean amount of electricity for their residents is more than the national average.
B. We fail to reject the null hypothesis and conclude that there is insufficient evidence at a 0.02 level of significance to support the power company claim that the mean amount of electricity for their residents is more than the national average.
C. We reject the null hypothesis and conclude that there is sufficient evidence at a 0.02 level of significance to support the power company's claim that the mean amount of electricity for their residents is more than the national average.
D. We fail to reject the null hypothesis and conclude that there is sufficient evidence at a 0.02 level of significance to support the power company's claim that the mean amount of electricity for their residents is more than the national average.
Mathematics
1 answer:
vladimir2022 [97]2 years ago
5 0

Answer:

Kindly check explanation for (1)

2.)

B. We fail to reject the null hypothesis and conclude that there is insufficient evidence at a 0.02 level of significance to support the power company claim that the mean amount of electricity for their residents is more than the national average.

Step-by-step explanation:

1.)

H0 : μ = 10290

H0 : μ > 10290

n = 153, sample mean, xbar = 10586 ; population standard deviation, σ = 2509 ; α = 0.05

Test statistic:

(xbar - μ) ÷ (σ/√(n))

(10586 - 10290) ÷ (2509/√(153))

296 / 202.84062

Test statistic (Z) = 1.46

We obtain the Pvalue :

Decision Region :

Reject H0 ; if Pvalue < α

Pvalue from Z test statistic calculator :

Pvalue = 0.072

Pvalue > α ; Hence, we fail to reject the Null ; there isn't sufficient evidence to support the power company's claim

2.)

H0 : μ = 10290

H0 : μ > 10290

n = 153, sample mean, xbar = 10446 ; population standard deviation, σ = 2509 ; α = 0.02

Test statistic:

(xbar - μ) ÷ (σ/√(n))

(10446 - 10290) ÷ (2509/√(153))

156 / 202.84062

Test statistic (Z) = 0.769

We obtain the Pvalue :

Decision Region :

Reject H0 ; if Pvalue < α

Pvalue from Z test statistic calculator :

Pvalue = 0.221

Pvalue > α ; Hence, we fail to reject the Null ;

B. We fail to reject the null hypothesis and conclude that there is insufficient evidence at a 0.02 level of significance to support the power company claim that the mean amount of electricity for their residents is more than the national average.

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Answer:

Kindly check explanation

Step-by-step explanation:

Given the data:

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Proportion who do think country is headed in the right direction :

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Margin of Error = Zcritical * √p(1-p)/n

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C.)

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D.)

Confidence interval for people who don't think :

P = 553 / 1007 = 0.549

Confidence interval = P ± margin of error

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Margin of Error = Zcritical * √p(1-p)/n

1-p = 1 - 0.549 = 0.451

Zcritical at 95% = 1.96

Margin of error = 1.96 * √0.549(0.451)/1007

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E.)

The interval with smaller error margin is ; 0.030 (those who think country is headed in the right direction because sample proportion is farther from .5 / farther from 1)

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Step-by-step explanation:

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