Question

In a random sample of 39 individuals, the average number of minutes spent exercising each week...

In a random sample of 39 individuals, the average number of minutes spent exercising each week was 124.7 minutes with a population standard deviation of 34.3 minutes. Find the 91% confidence interval for the mean. Show your work to receive credit

Homework Answers

Answer #1

Note : Here, t-critical value is used.

If z -critical value is used the confidence interval is (115.363 , ​​​​​​134.037)

final answers are rounded to 3 decimals

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
In a random sample of 39 individuals the average number of minutes spent exercising in each...
In a random sample of 39 individuals the average number of minutes spent exercising in each week what is 124.7 minutes with a population standard deviation of 34.3 minutes.Find the 91% confidence interval for the mean
in a random sample of 24 people the average number of minutes spent sleeping each day...
in a random sample of 24 people the average number of minutes spent sleeping each day is 434.5 minutes with a sample standard deviation of 77.2 minutes. Find the 90% confidence interval for the mean
A study regarding those that drink coffee showed that in a random sample of 18 individuals,...
A study regarding those that drink coffee showed that in a random sample of 18 individuals, the standard deviation for the number of times an individual drank coffee during a week was 5.2. Find the 95% confidence interval of the variance and the standard deviation. Show your work to receive credit
In a sample of 49 individuals, it is found that the average amount spent on groceries...
In a sample of 49 individuals, it is found that the average amount spent on groceries per month is $255. From previous studies, it is assumed that the population standard deviation is $46. Find a 95% confidence interval for the average amount of money spent on groceries per month, and interpret your result. Use ??=1.96 for 95% confidence, and round your answers to 2 decimal places. The confidence interval is: $ to $ This means: Select your answer from one...
In a sample of 40 individuals, it is found that the average amount spent on groceries...
In a sample of 40 individuals, it is found that the average amount spent on groceries per month is $255. From previous studies, it is assumed that the population standard deviation is $46. Find a 95% confidence interval for the average amount of money spent on groceries per month, and interpret your result. Use ??=1.96 for 95% confidence, and round your answers to 2 decimal places. The confidence interval is: $________ to $________ This means: Select your answer from one...
A sample of 39 individuals at a shopping mall found that the mean of visits to...
A sample of 39 individuals at a shopping mall found that the mean of visits to a restaurant per week was 2.88 with a standard deviation of 1.53. Find a 90% confidence interval for the mean number of restaurants visits.
A random sample of 90 UT business students revealed that on average they spent 8 hours...
A random sample of 90 UT business students revealed that on average they spent 8 hours per week on Facebook. The population standard deviation is assumed to be 2 hours per week. If we want to develop a 95% confidence interval for the average time spent per week on Facebook by all UT business students, the margin of error of this interval is The 95% confidence interval for the average time spent per week on Facebook by all UT business...
In a sample of 10 CEOs, they spent an average of 12.9 hours each week looking...
In a sample of 10 CEOs, they spent an average of 12.9 hours each week looking into new product opportunities with a sample standard deviation of 4.9 hours. Find the 95% confidence interval. Assume the times are normally distributed. (8.0, 17.8) (9.9, 15.9) (9.4, 16.4) (11.1, 14.7)
A random sample found that tourists in Chicago spent an average of $2860 per week, with...
A random sample found that tourists in Chicago spent an average of $2860 per week, with a standard deviation of $126. A separate random sample found that tourists in Miami spent an average of $2935 per week, with a standard deviation of $138. The sample sizes are 89 for Chicago and 64 for Miami. Suppose that you want to find out if there is any difference between the average expenditures of all of the tourists who visit the two cities....
A random sample of 100 workers in one large plant took an average of 12 minutes...
A random sample of 100 workers in one large plant took an average of 12 minutes to complete a task, with a standard deviation of 2 minutes. A random sample of 50 workers in a second large plant took an average of 11 minutes to complete the task, with a standard deviation of 3 minutes. Construct a 95% confidence interval for the difference between the two population mean completion times.
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT