Question

You are constructing a marketing research study to determine what proportion of consumers between the ages...

You are constructing a marketing research study to determine what proportion of consumers between the ages of 18 and 35 own DVD players. You have randomly sample 60 individuals in that age range, and you have determined that 11 of them own DVD players.

a.Find the 90% confidence interval for the proportion.

b. Suppose that your contract for the study requires that you estimate the proportion within plus or minus 0.05 with 90% confidence. How many additional subjects would you need to sample to obtain that level of precision?

Homework Answers

Answer #1

a)

Sample proportion = 11 / 60 = 0.183

Margin of error at 90% confidence = Z/2 * sqrt [ p (1 - p) / n]

= 1.645 * sqrt [ 0.183 ( 1 - 0.183 ) / 60 ]

= 0.082

90% confidence interval is

- E < p < + E

0.183 - 0.082 < p < 0.183 + 0.082

0.101 < p < 0.265

90% CI is ( 0.101 , 0.265 )

b)

Sample szie = Z2/2 * p ( 1 - p) / E2

= 1.64492 * 0.183 ( 1 - 0.183) / 0.052

= 162.01

sample size = 163 (Rounded up to nearest integer)

Additional sample required = 163 - 60 = 103

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
1. When constructing a confidence interval to estimate a population proportion, what affects the size of...
1. When constructing a confidence interval to estimate a population proportion, what affects the size of the margin of error? A. The sample size B. The sample proportion C. The confidence level D. All of the above affect the size of the margin of error E. None of the above affect the size of the margin of error 2. What percentage of couples meet through online dating apps? A survey of a random sample of couples finds that 12% say...
A political candidate has asked you to conduct a poll to determine what percentage of people...
A political candidate has asked you to conduct a poll to determine what percentage of people support him. If the candidate only wants a 4% margin of error at a 97.5% confidence level, what size of sample is needed? When finding the z-value, round it to four decimal places. You want to obtain a sample to estimate a population proportion. At this point in time, you have no reasonable preliminary estimation for the population proportion. You would like to be...
****PLEASE ANSWER ALL OF FOUR QUESTIONS!!!!**** QUESTION 1: You are curious about the average number of...
****PLEASE ANSWER ALL OF FOUR QUESTIONS!!!!**** QUESTION 1: You are curious about the average number of yards Matthew Stafford throws for each game for the Detroit Lions. You randomly select 30 games and see that the average yards per game is 269.1 with a standard deviation of 27.31 yards. You want to create a 90% confidence interval for the true average number of yards per game he throws. What is the margin of error for this estimate? QUESTION 2: You...
In a study of health behavior, a random sample of 500 subjects from a major city...
In a study of health behavior, a random sample of 500 subjects from a major city was surveyed. In this random sample, 310 responded that they had dinner mostly after 8 PM. At 5% level of significance, does the data support that more than 60% of the residents in this city had dinner after 8 PM? (Must state null and alternative hypotheses and use p-value to conclude the test. You can use Chi-square Goodness of fit test or Binomial test...
In a study of health behavior, a random sample of 500 subjects from a major city...
In a study of health behavior, a random sample of 500 subjects from a major city was surveyed. In this random sample, 310 responded that they had dinner mostly after 8 PM. At 5% level of significance, does the data support that more than 60% of the residents in this city had dinner after 8 PM? (Must state null and alternative hypotheses and use p-value to conclude the test. You can use Chi-square Goodness of fit test or Binomial test...
Recall in our discussion of the binomial distribution the research study that examined schoolchildren developing nausea...
Recall in our discussion of the binomial distribution the research study that examined schoolchildren developing nausea and vomiting following holiday parties. The intent of this study was to calculate probabilities corresponding to a specified number of children becoming sick out of a given sample size. Recall also that the probability, i.e. the binomial parameter "p" defined as the probability of "success" for any individual, of a randomly selected schoolchild becoming sick was given. Suppose you are now in a different...
1. To estimate the standard error of the mean, a survey researcher will: Select one: a....
1. To estimate the standard error of the mean, a survey researcher will: Select one: a. take the square root of the sample standard deviation. b. none of the above. c. multiply the sample standard deviation times the square root of the sample size. d. divide the sample standard deviation by the square root of sample size. 2. Suppose that a business researcher is attempting to estimate the amount that members of a target market segment spend annually on detergent...
Question 5: You work for the consumer insights department of a major big box retailer and...
Question 5: You work for the consumer insights department of a major big box retailer and you are investigating the efficacy of a new e-mail marketing campaign. Through the use of e-mail analytics research, you have determined that in a sample of 981 monitored subscribers, 289 of them opened the e-mail within 24 hours of receiving it. What is the 95% confidence interval for the true proportion of all e-mail subscribers that opened the e-mail within 24 hours of receiving...
You own a small storefront retail business and are interested in determining the average amount of...
You own a small storefront retail business and are interested in determining the average amount of money a typical customer spends per visit to your store. You take a random sample over the course of a month for 45 customers and find that the average dollar amount spent per transaction per customer is $90.422 with a standard deviation of $13.895. When creating a 90% confidence interval for the true average dollar amount spend per customer, what is the margin of...
Recall in our discussion of the normal distribution the research study that examined the blood vitamin...
Recall in our discussion of the normal distribution the research study that examined the blood vitamin D levels of the entire US population of landscape gardeners. The intent of this large-scale and comprehensive study was to characterize fully this population of landscapers as normally distributed with a corresponding population mean and standard deviation, which were determined from the data collection of the entire population. Suppose you are now in a different reality in which this study never took place though...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT