Question

A developer is considering two alternative sites for a regional shopping center. Since household income in...

A developer is considering two alternative sites for a regional shopping center. Since household income in the community is one important consideration in such site selection, the developer wishes to test the null hypothesis that there is no difference between the mean household income amounts in the two communities. Consistent with this hypothesis, it is assumed that the standard deviation of household income is also the same in the two communities. For a sample of 29 households in the first community, the average annual income is $44,500 with the sample standard deviation $1,700. For a sample of 40 households in the second community, average annual income is $43,200 and standard deviation is $2,250. Test the null hypothesis at the 5 percent level of significance.

Homework Answers

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
A representative of a community group informs the prospective developer of a shopping center that the...
A representative of a community group informs the prospective developer of a shopping center that the average income per household in the area is $45 000. Suppose that for the type of area the observed household income can be assumed to be approximately normally distributed, and that the standard deviation can be accepted as being equal to $2000, based on an earlier study. In a random sample of 15 households, the mean household income is found to be $44 000....
In 2016, the mean household income in Dallas was $39,097. A researcher thinks that household income...
In 2016, the mean household income in Dallas was $39,097. A researcher thinks that household income has risen since then, and that the mean household income in Dallas is higher than $39,097. She gathers income data from a random sample of 150 Dallas households. The sample has a mean household income of $40,352, and a standard deviation of 1,875. Based on this information, test the researcher's hypothesis at a significance level of .01. Be sure to include all 4 steps.
According to the Census Bureau, 3.39 people reside in the typical American household. A sample of...
According to the Census Bureau, 3.39 people reside in the typical American household. A sample of 26 households in Arizona retirement communities showed the mean number of residents per household was 2.73 residents. The standard deviation of this sample was 1.22 residents. At the .10 significance level, is it reasonable to conclude the mean number of residents in the retirement community household is less than 3.39 persons? ate the null hypothesis and the alternate hypothesis. (Round your answer to 2...
According to the Census Bureau, 3.34 people reside in the typical American household. A sample of...
According to the Census Bureau, 3.34 people reside in the typical American household. A sample of 27 households in Arizona retirement communities showed the mean number of residents per household was 2.86 residents. The standard deviation of this sample was 1.13 residents. At the .05 significance level, is it reasonable to conclude the mean number of residents in the retirement community household is less than 3.34 persons? a. State the null hypothesis and the alternate hypothesis. (Round your answer to...
An advertising executive claims that there is a difference in the mean household income for credit...
An advertising executive claims that there is a difference in the mean household income for credit cardholders of Visa Gold and of MasterCard Gold. A random survey of 18 Visa Gold cardholders resulted in a mean household income of $84,280 with a standard deviation of $10,400. A random survey of 11 MasterCard Gold cardholders resulted in a mean household income of $79,360 with a standard deviation of $9800. Is there enough evidence to support the executive's claim? Let μ1 be...
According to the Census Bureau, 3.26 people reside in the typical American household. A sample of...
According to the Census Bureau, 3.26 people reside in the typical American household. A sample of 26 households in Arizona retirement communities showed the mean number of residents per household was 2.89 residents. The standard deviation of this sample was 1.25 residents. At the .05 significance level, is it reasonable to conclude the mean number of residents in the retirement community household is less than 3.26 persons? (a) State the null hypothesis and the alternate hypothesis. (Round your answer to...
Identify the P-VALUE in a hypothesis test of the following claim and sample data: Claim: "The...
Identify the P-VALUE in a hypothesis test of the following claim and sample data: Claim: "The average annual household income in Warren County is $47,500." A random sample of 86 households from this county is obtained, and they have an average annual income of $48,061 with a standard deviation of $2,351. Test the claim at the 0.02 significance level. a. 0.013 b. 0.015 c. 0.030 d. 0.027 Identify the CONCLUSION of a hypothesis test of the following claim and sample...
A politician claims the mean household income in Indiana is $60,000. A reporter believes it is...
A politician claims the mean household income in Indiana is $60,000. A reporter believes it is less than $60,000. The reporter selects a random sample of 50 Indiana households. The mean income computed from the 50 people surveyed is $56,000. (Assume the standard deviation in Indiana household income is $15,000.) We wish to do a hypothesis test to determine whether this provides evidence (at the .05 level) that the mean household income is less than $60,000. You should complete this...
An advertising executive claims that there is a difference in the mean household income for credit...
An advertising executive claims that there is a difference in the mean household income for credit cardholders of Visa Gold and of MasterCard Gold. A random survey of 13 Visa Gold cardholders resulted in a mean household income of $68,600 with a standard deviation of $9200. A random survey of 8 MasterCard Gold cardholders resulted in a mean household income of $59,620 with a standard deviation of $10,900. Is there enough evidence to support the executive's claim? Let μ1 be...
An advertising executive claims that there is a difference in the mean household income for credit...
An advertising executive claims that there is a difference in the mean household income for credit cardholders of Visa Gold and of MasterCard Gold. A random survey of 17 Visa Gold cardholders resulted in a mean household income of $⁢58,390 with a standard deviation of $⁢8500. A random survey of 11 MasterCard Gold cardholders resulted in a mean household income of $⁢54,320 with a standard deviation of $⁢10,500. Is there enough evidence to support the executive's claim? Let μ1 be...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT