Question

In a recent poll of 210 south Jersey households, it was found that 152 households had...

In a recent poll of 210 south Jersey households, it was found that 152 households had at least one computer. Construct a 95% confidence interval to estimate the population proportion of south Jersey. A separate survey in Philadelphia found the sample proportion of households with more than one computer to be 0.66, what kind conclusion you can made in terms of the difference in proportion between South Jersey and Philadelphia? (Your conclusion must be based on the confidence interval you obtained.

Math   Statistics-And-Probability   
0 0
Add a commentTranscribed image text
Next >

Homework Answers

Answer #1

Sample proportion,   = 152/210 =  0.7238

Standard error of sample proportion, SE = = 0.030854

z score for 95% confidence interval is 1.96

95% confidence interval for population proportion of south Jersey households with at least one computer is,

0.7238 - 1.96 * 0.030854 to 0.7238 + 1.96 * 0.030854

0.6633 to  0.7843

Since the 95% confidence interval for population proportion of south Jersey households with at least one computer does not contain the value 0.66, we conclude that there is significant difference in proportion of households with at least one computer between South Jersey and Philadelphia.

0 0
Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
In a recent poll of 210 south Jersey households, it was found that 152 households had...
In a recent poll of 210 south Jersey households, it was found that 152 households had at least one computer. Construct a 95% confidence interval to estimate the population proportion of south Jersey. A separate survey in Philadelphia found the sample proportion of households with more than one computer to be 0.66, what kind conclusion you can made in terms of the difference in proportion between South Jersey and Philadelphia? (Your conclusion must be based on the confidence interval you...
In a recent poll of 200 households, it was found that 152 had at least one...
In a recent poll of 200 households, it was found that 152 had at least one computer and one television. A 95% confidence interval to estimate the population proportion was calculated to be .701 to .819. What sample size would be needed to change the margin of error to .030?
4.1       A recent survey found that 35 people out of a sample of 410 South Africans...
4.1       A recent survey found that 35 people out of a sample of 410 South Africans suffer from diabetes. Construct a 95% confidence interval for the true proportion of all South Africans who suffer from diabetes. Interpret your answer. (6) Confidence Interval Confidence Level Z - Limits 90% ± 1.645 95% ±1.96 99% ±2.58
In a recent survey of 4276 randomly selected households showed that 4020 of them had telephones....
In a recent survey of 4276 randomly selected households showed that 4020 of them had telephones. Using these results, construct a 99% confidence interval estimate of the true proportion of households with telephones. P – PARAMETER Describe the parameter in context. This should be a complete sentence A – ASSUMPTIONS Are we allowed to construct an interval and why? N – NAME THE INTERVAL State what command you will be using the compute the interval along with the values you...
A recent survey of 200 households showed that 8 had a single male as the head...
A recent survey of 200 households showed that 8 had a single male as the head of household. Fortyyears ago, a survey of 200 households showed that 6 had a single male as the head of household. Canit be concluded that the proportion has changed? Find the 95% confidence interval of the differenceof the two proportions. Does the confidence interval contain zero? Why is this important to know? For all confidence intervals, include a. difference between sample estimates b. standard...
In a recent poll of 200 households in the area. 122 households had two or more...
In a recent poll of 200 households in the area. 122 households had two or more vehicles. Using a 99% level of confidence, estimate the true proportion of households in the area that have two or more vehicles. What is the correct critical value to be used in calculating the interval estimate? a. 1.645 b. 1.414 c. 2.575 d. 2.018 What is the 99% confidence interval estimate? a. (0.521, 0.699) b. (0.542, 0.678) c. (0.553, 0.667) d. (0.513, 0.707) With...
A recent survey of 200 households showed that 8 had a single male as the head...
A recent survey of 200 households showed that 8 had a single male as the head of the household. Forty years ago, a survey of 200 households showed that 6 had a single male as the head of the household. Can it be concluded that the proportion has changed, based on a 95% confidence interval? Recall the confidence interval is calculated as
A recent poll of 120 adults who frequent the local farmer’s market found that 54 have...
A recent poll of 120 adults who frequent the local farmer’s market found that 54 have purchased reusable cloth bags for their groceries. Based on these data, construct and interpret a 95% confidence interval of the proportion of adults who have purchased reusable cloth bags? 1.Type the lower and upper bounds of the confidence interval. 2.Type the interpretation of the confidence interval.
Suppose the marketing company did do a survey. They randomly surveyed 200 households and found that...
Suppose the marketing company did do a survey. They randomly surveyed 200 households and found that in 120 of them, the woman made the majority of the purchasing decisions. We are interested in the population proportion of households where women make the majority of the purchasing Construct a 95% confidence interval for the population proportion of households where the women make the majority of the purchasing decisions. State the confidence interval, sketch the graph and calculate the error bound.
A national survey of 47,000 American households in 2006 found that 32.4% of the households included...
A national survey of 47,000 American households in 2006 found that 32.4% of the households included a pet cat. No information was provided about how the sample households were selected. Who are the subjects? what is the variable? Is the variable quantitative or categorical? If the standard deviation of the null distribution is 0.011, what is the 95% confidence interval using the 2SD method? Based on the confidence interval you found, do you believe that over one-third of American households...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT