A company claims that the mean monthly residential electricity consumption in a certain region is more than 890 kiloWatt-hours (kWh). You want to test this claim. You find that a random sample of 69 residential customers has a mean monthly consumption of 910 kWh. Assume the population standard deviation is 122 kWh. At α=0.05, can you support the claim? Complete parts (a) through (e). (a) Identify H0 and Ha. Choose the correct answer below. A. H0: μ>910 (claim) Ha: μ≤910 B. H0: μ=910 Ha: μ≠910 (claim) C. H0: μ≤910 Ha: μ>910 (claim) D. H0: μ>890 (claim) Ha: μ≤890 E. H0: μ≤890 Ha: μ>890 (claim) F. H0: μ=890 (claim) Ha: μ≠890 (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 value is nothing. B. The critical values are ±nothing. Identify the rejection region(s). Select the correct choice below. A. The rejection region is z<1.64. B. The rejection regions are z<−1.64 and z>1.64. C. The rejection region is z>1.64. (c) Find the standardized test statistic. Use technology. The standardized test statistic is z=nothing. (Round to two decimal places as needed.) (d) Decide whether to reject or fail to reject the null hypothesis. A. Fail to reject H0 because the standardized test statistic is not in the rejection region. B. Reject H0 because the standardized test statistic is in the rejection region. C. Fail to reject H0 because the standardized test statistic is in the rejection region. D. Reject H0 because the standardized test statistic is not in the rejection region. (e) Interpret the decision in the context of the original claim. At the 5% significance level, there ▼ is notis not isis enough evidence to ▼ support reject the claim that the mean monthly residential electricity consumption in a certain region ▼ is different from is greater than is less than nothing kWh.

A company claims that the mean monthly residential electricity consumption in a certain region is more than 890 kiloWatt-hours (kWh). You want to test this claim. You find that a random sample of 69 residential customers has a mean monthly consumption of 910 kWh. Assume the population standard deviation is 122 kWh. At α=0.05, can you support the claim? Complete parts (a) through (e). (a) Identify H0 and Ha. Choose the correct answer below. A. H0: μ>910 (claim) Ha: μ≤910 B. H0: μ=910 Ha: μ≠910 (claim) C. H0: μ≤910 Ha: μ>910 (claim) D. H0: μ>890 (claim) Ha: μ≤890 E. H0: μ≤890 Ha: μ>890 (claim) F. H0: μ=890 (claim) Ha: μ≠890 (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 value is nothing. B. The critical values are ±nothing. Identify the rejection region(s). Select the correct choice below. A. The rejection region is z<1.64. B. The rejection regions are z<−1.64 and z>1.64. C. The rejection region is z>1.64. (c) Find the standardized test statistic. Use technology. The standardized test statistic is z=nothing. (Round to two decimal places as needed.) (d) Decide whether to reject or fail to reject the null hypothesis. A. Fail to reject H0 because the standardized test statistic is not in the rejection region. B. Reject H0 because the standardized test statistic is in the rejection region. C. Fail to reject H0 because the standardized test statistic is in the rejection region. D. Reject H0 because the standardized test statistic is not in the rejection region. (e) Interpret the decision in the context of the original claim. At the 5% significance level, there ▼ is notis not isis enough evidence to ▼ support reject the claim that the mean monthly residential electricity consumption in a certain region ▼ is different from is greater than is less than nothing kWh.

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Question

A company claims that the mean monthly residential electricity consumption in a certain region is more than

890

kiloWatt-hours (kWh). You want to test this claim. You find that a random sample of

residential customers has a mean monthly consumption of

910

kWh. Assume the population standard deviation is

122

kWh. At

α=0.05,

can you support the claim? Complete parts (a) through (e).

(a) Identify

and

Ha.

Choose the correct answer below.

H0:

μ>910

(claim)

Ha:

μ≤910

H0:

μ=910

Ha:

μ≠910

(claim)

H0:

μ≤910

Ha:

μ>910

(claim)

H0:

μ>890

(claim)

Ha:

μ≤890

H0:

μ≤890

Ha:

μ>890

(claim)

H0:

μ=890

(claim)

Ha:

μ≠890

(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

nothing.

The critical values are

±nothing.

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

The rejection region is

z<1.64.

The rejection regions are

z<−1.64

and

z>1.64.

The rejection region is

z>1.64.

The standardized test statistic is

z=nothing.

(Round to two decimal places as needed.)

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

Fail to reject

because the standardized test statistic

is not

in the rejection region.

Reject

because the standardized test statistic

in the rejection region.

Fail to reject

because the standardized test statistic

in the rejection region.

Reject

because the standardized test statistic

is not

in the rejection region.

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

At the

significance level, there

▼

is notis not

isis

enough evidence to

▼

support

reject

the claim that the mean monthly residential electricity consumption in a certain region

▼

is different from

is greater than

is less than

nothing

kWh.

Click to select your answer(s).

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

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