Question

The growing season for a random sample of 25 U.S. cities were recorded, yielding a sample...

The growing season for a random sample of 25 U.S. cities were recorded, yielding a sample mean of 190.7 days and the sample standard deviation of 54.2 days.

a) Find the 95% confidence interval of the population mean of the days of the growing season.

b) Interpret the confidence interval from part a)

Homework Answers

Answer #1

Confidence interval calculation.

Here population standard deviation is unknown so, t test confidence interval calculation.

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
Exercise 1Pulse rates were measured on a random sample of 25 distance runners. The dataresulted in...
Exercise 1Pulse rates were measured on a random sample of 25 distance runners. The dataresulted in an average pulse rate of ̄x= 58.2 beats per minute with a standarddeviation ofs= 8.5. (a) Find a 95% confidence interval for the mean pulse rate for the population ofdistance runners. (b) Interpret the interval in words.
A simple random sample of 25 items from a normally distributed population resulted in a sample...
A simple random sample of 25 items from a normally distributed population resulted in a sample mean of 80 and a sample standard deviation of 7.5. Construct a 95% confidence interval for the population mean.
The rising cost of malpractice insurance is a growing problem. A random sample of thirty-three General...
The rising cost of malpractice insurance is a growing problem. A random sample of thirty-three General Surgery claims from TN had a mean cost of $47505 with a standard deviation of $1535. a. Find a 95% confidence interval for the mean claim amount of all General Surgery claims in TN. b. Interpret the confidence interval in the words of the problem. c. Find the error bound. d. A senator claims that mean for all General Surgery claims in TN is...
The rising cost of malpractice insurance is a growing problem. A random sample of thirty-three General...
The rising cost of malpractice insurance is a growing problem. A random sample of thirty-three General Surgery claims from TN had a mean cost of $47505 with a standard deviation of $1535. a. Find a 95% confidence interval for the mean claim amount of all General Surgery claims in TN. b. Interpret the confidence interval in the words of the problem. c. Find the error bound. d. A senator claims that mean for all General Surgery claims in TN is...
In a random sample of 25 ​people, the mean commute time to work was 33.2 minutes...
In a random sample of 25 ​people, the mean commute time to work was 33.2 minutes and the standard deviation was 7.1 minutes. Assume the population is normally distributed and use a​ t-distribution to construct a 95​% confidence interval for the population mean muμ. What is the margin of error of muμ​? Interpret the results. The confidence interval for the population mean muμ is (____,_______ ) ​(Round to one decimal place as​ needed.) The margin of error of muμ is...
In a random sample of 25 students, it was found that they consumed on average 23...
In a random sample of 25 students, it was found that they consumed on average 23 grams of sugar each day with a standard deviation of 4 grams. a) Construct a 99% confidence interval for the mean amount of sugar consumed by students. b) Interpret the confidence interval found in part a.
. The winning team's scores in 13 high school basketball games were recorded. If the sample...
. The winning team's scores in 13 high school basketball games were recorded. If the sample gave a mean of 10.5 points with a sample standard deviation of 0.25 points, find the 98% confidence interval of the true mean. Interpret the confidence interval. Point Estimate: ______ Critical Value: Margin of Error: ________ Confidence Interval: __________________________________________________ Interpretation: _______________________________________________
Consider two populations. A random sample of 28 observations from the first population revealed a sample...
Consider two populations. A random sample of 28 observations from the first population revealed a sample mean of 40 and a sample standard deviation of 12. A random sample of 32 observations from the second population revealed a sample mean of 35 and a sample standard deviation of 14. (a) Using a .05 level of significance, test the hypotheses H0 : μ1 − μ2 = 0 and H1 : μ1 − μ2 ≠ 0 respectively. Explain your conclusions. (b) What...
In a random sample of six mobile​ devices, the mean repair cost was $75.00 and the...
In a random sample of six mobile​ devices, the mean repair cost was $75.00 and the standard deviation was $13.0013.00. Assume the population is normally distributed and use a​ t-distribution to find the margin of error and construct a 95​% confidence interval for the population mean. Interpret the results. The 95​% confidence interval for the population mean μ is
A large hospital wholesaler, as part of an assessment of workplace safety, gave a random sample...
A large hospital wholesaler, as part of an assessment of workplace safety, gave a random sample of 54 of its warehouse employees a test (measured on a 0 to 100 point scale) on safety procedures. For that sample of employees, the mean test score was 75 points, with a sample standard deviation of 15 points. Determine and interpret a 95% confidence interval for the mean test score of all the company’s warehouse employees. (Please keep at least four decimal places)....
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT