Question

A survey of 90 randomly selected homeowners finds that they spend a mean of $66 per...

A survey of 90 randomly selected homeowners finds that they spend a mean of $66 per month on home maintenance. Assume that the population standard deviation is $13 per month. Construct and interpret a 95% confidence interval for the mean amount of money spent per month on home maintenance by all homeowners.

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

Homework Answers

Answer #1

Z for 95% confidence interval = Z0.025 = 1.96

Confidence interval

                              

                              

                               = (63.314 , 68.686)

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
A survey of 87 randomly selected homeowners finds that they spend a mean of $65 per...
A survey of 87 randomly selected homeowners finds that they spend a mean of $65 per month on home maintenance. Assume that the population standard deviation is $13 per month. Construct and interpret a 97% confidence interval for the mean amount of money spent per month on home maintenance by all homeowners.
A survey of 100 randomly selected Michigan homeowners finds that they spend a mean of $67...
A survey of 100 randomly selected Michigan homeowners finds that they spend a mean of $67 per month on home maintenance. In an independent survey, 100 randomly selected Florida homeowners find that they spend a mean of $75 per month on home maintenance. Construct a 99% confidence interval for the true difference in the mean amounts of money spent per month on home maintenance for the two states. Assume that the population standard deviation is $14 per month for Michigan...
A survey of 185 randomly selected homeowners found that they spend a mean of $67 per...
A survey of 185 randomly selected homeowners found that they spend a mean of $67 per month on home maintenance, with a standard deviation of $14 per month. Estimate the mean amount of money that a homeowner can expect to spend monthly on maintenance at a 95% confidence level. You do not need to check conditions. Be sure to interpret your answer in a sentence.
A survey of 100 randomly selected Michigan residents finds that the they spend a mean of...
A survey of 100 randomly selected Michigan residents finds that the they spend a mean of $21 per month on going to the movies. An independent study of a 100 randomly selected California residents find that they spend a mean of $46 per month on going to the movies. Assume that the population standard deviation for the Michigan residents is $5 per month and that the population standard deviation for the California residents is $23 per month. Answer the following...
A survey of 20 randomly selected adult men showed that the mean time they spend per...
A survey of 20 randomly selected adult men showed that the mean time they spend per week watching sports on television is 9.34 hours with a standard deviation of 1.34 hours. Assuming that the time spent per week watching sports on television by all adult men is (approximately) normally distributed, construct a 90 % confidence interval for the population mean,   μ . Round your answers to two decimal places. Lower bound: Enter your answer; confidence interval, lower bound Upper bound:...
25 randomly selected students took a survey on how long they spent on social media. The...
25 randomly selected students took a survey on how long they spent on social media. The mean number of minutes these students spent on social media was 205.1 minutes with a standard deviation of 150 minutes.The population of the time spent on social media is normally distributed. Construct a 98%confidence interval for the population standard deviation of the time students spend on social media.
In a survey of randomly selected subjects, the mean age of 40 respondents is 36.4 years,...
In a survey of randomly selected subjects, the mean age of 40 respondents is 36.4 years, and the sample standard deviation of all ages is known to be 10.1 years. Construct a 95% confidence interval estimate of the mean age of the population from which the sample was selected.
A random sample of 58 randomly selected people were asked how much they spend on their...
A random sample of 58 randomly selected people were asked how much they spend on their grocery bill each week. The mean was $115.36 with a standard deviation of $16.6. Find a 95% confidence interval for the true population mean amount spent on groceries each week. Report the answer accurate to four decimal places.
In a random sample of 47 high-school students enrolled in SAT test training, it was determined...
In a random sample of 47 high-school students enrolled in SAT test training, it was determined that the mean amount of money spent per student was $110.96 per class with a sample standard deviation of $8.30. Construct and interpret a 95% confidence interval for the population mean amount spent per student for an SAT class. A) ($108.52, $113.40); We are 95% confident that the population mean amount spent per student for a single SAT class is between $108.52 and $113.40....
A survey of 25 randomly selected customers had the mean 33.04 years and the standard deviation...
A survey of 25 randomly selected customers had the mean 33.04 years and the standard deviation 10.45 years. What is the standard error of the mean? How would the standard error change if the sample size had been             400 instead of 25? (Assume that the sample standard deviation didn't change.) Find the degrees of freedom for n=60. Find the t critical value for n=60 for 90% CI for mean. A survey of 25 randomly selected customers had the mean...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT