Question

A random sample of 40 adults aged 25 to 34 years spent an average of $216...

A random sample of 40 adults aged 25 to 34 years spent an average of $216 annually on movies The sample standard deviation is $60.

a) Develop a 95% confidence interval for the average annual amount spent on movies.  

b) Verbally interpret the meaning of the confidence interval that you found for part a

c) If the desired margin of error for the above confidence interval is $10, what size sample is necessary? using the sample standard deviation given above as the planning value for the population standard deviation

0 0
Add a comment Improve this question Transcribed image text
Know the answer?
Add Answer to:
A random sample of 40 adults aged 25 to 34 years spent an average of $216...
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. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • 1. A random sample of 37 employees in Sweden worked an average number of 231 days...

    1. A random sample of 37 employees in Sweden worked an average number of 231 days per year. The sample standard deviation is 26 days. a) Develop a 95% confidence interval for the average number of work-days per year for all Swedish employees. (6 marks) b) Verbally interpret the meaning of the confidence interval that you found for part a) c) If the desired margin of error for the above confidence interval is 4 days, what size sample is necessary?...

  • 1. How many movies does an adult watch on average each year? A random sample of...

    1. How many movies does an adult watch on average each year? A random sample of 29 adults with an average of 128.6 movies and standard deviation of 23.8 movies per year. Construct 95% confidence interval for the average number of movies an adult would watch each year. a. Find the critical value. b. Calculate the margin of error Write the interval. c.

  • The fitness of 514 adults (aged 25 to 35) were sampled and their heart rates measured...

    The fitness of 514 adults (aged 25 to 35) were sampled and their heart rates measured immediately after exercise. This sample of adults was found to have a mean of 1427 beats (per 10 minutes) with a standard deviation of 325 beats (per 10 minutes). Construct a two-tailed 95% confidence interval for the true mean heart beat of all such adults. Determine the sample size that would be necessary to obtain an interval with width (at most) 50 for a...

  • 1. A random sample of 82 customers, who visited a department store, spent an average of...

    1. A random sample of 82 customers, who visited a department store, spent an average of $71 at this store. Suppose the standard deviation of expenditures at this store is O = $19. What is the e 98% confidence interval for the population mean? 2. A sample of 25 elements produced a mean of 123.4 and a standard deviation of 18.32 Assuming that the population has a normal distribution, what is the 90% confidence interval for the population mean? 3....

  • The average selling price of a smartphone purchased by a random sample of 43 customers was...

    The average selling price of a smartphone purchased by a random sample of 43 customers was ​$315. Assume the population standard deviation was ​$34. a. Construct a 95​% confidence interval to estimate the average selling price in the population with this sample. b. What is the margin of error for this​ interval? a. The 95​% confidence interval has a lower limit of and an upper limit of. ​(Round to the nearest cent as​ needed.) b. The margin of error is....

  • Was the A random sample of 40 adults with no children under the age of 18...

    Was the A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.96 hours, with a standard deviation of 226 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4 23 hours, with a standard deviation of 155 hours. Construct and interpret a 95% confidence interval for the mean difference in leisure time between...

  • A random sample of 40 adults with no children under the age of 18 years results...

    A random sample of 40 adults with no children under the age of 18 years results in a meandaly leisure time of 5.92 hours, with a standard deviation of 2.33 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.26 hours, with a standard deviation of 1.61 hours. Construct and interpret a 95% confidence interval for the mean difference in leisure time between adults with no children...

  • A sample of 100 consumers shows that the average amount spent annually on a debit card...

    A sample of 100 consumers shows that the average amount spent annually on a debit card is $8000. And assume that the population standard deviation of debit card expenditure is $500. Construct a 95% confidence interval of the population mean amount spent annually on a debit card.

  • A random sample of 40 adults with no children under the age of 18 years results...

    A random sample of 40 adults with no children under the age of 18 years results in a meandaly leisure time of 522 hours, with a standard deviation of 2.34 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.37 hours, with a standard deviation of 1.52 hours. Construct and interpreta 95% confidence interval for the mean difference in leisure time belwoon adults with no children and...

  • A survey of 25 randomly selected customers found that their average age was 31.84 years with...

    A survey of 25 randomly selected customers found that their average age was 31.84 years with a standard deviation of 9.84 years. What would the critical value t* be for a 95% confidence interval with 99 degrees of freedom? What would the margin of error be for a 95% confidence interval for this data (with the sample size of 25)? Construct a 95% confidence interval for the mean age of all customers for this data (with the sample size of...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT