Question

Suppose the Annual rainfall (in inches) in 5 different locations in Hawaii are: 12, 21, 14,...

Suppose the Annual rainfall (in inches) in 5 different locations in Hawaii are: 12, 21, 14, 16, 17. A statistician wants to use this data to test the claim that the average rainfall in Hawaii is more than 18 inches. Assume significance level α = 0.05 .   

Calculate the 90% Confidence Interval for the average rainfall in Hawaii, show excel commands and formulas used.

Homework Answers

Answer #1

Solution:

Confidence interval for Population mean is given as below:

Confidence interval = Xbar ± t*S/sqrt(n)

From given data, we have

Xbar = 16

S = 3.391164992

n = 5

df = n – 1 = 4

Confidence level = 90%

Critical t value = 2.1318

[by using excel command =TINV(1 - 0.90,4)]

Confidence interval = Xbar ± t*S/sqrt(n)

Confidence interval = 16 ± 2.1318*3.391164992/sqrt(5)

Confidence interval = 16 ± 3.2331

Lower limit = 16 - 3.2331 = 12.77

Upper limit = 16 + 3.2331 = 19.23

Confidence interval = (12.77, 19.23)

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
Suppose the Annual rainfall (in inches) in 5 different locations in Hawaii are: 28, 18, 18,...
Suppose the Annual rainfall (in inches) in 5 different locations in Hawaii are: 28, 18, 18, 16, 14. A statistician wants to use this data to test the claim that the average rainfall in Hawaii is more than 16 inches. Assume significance level α=0.05 .    a. What is the alternative hypothesis H1? What tailed test is this? b. Determine the value of the Test Statistic. c. What is the P-value? d. Should the statistician Reject or Fail to Reject...
A random sample of 35 locations in a western state yield an average annual precipitation of...
A random sample of 35 locations in a western state yield an average annual precipitation of 16.7 inches for last year. Suppose the standard deviation is 2.1 inches. If a meteorologist wants to estimate the rainfall mean within .2 inch with 90% confidence, how large of a sample does she need to take?
Suppose we have the following data 12 17 13 25 16 21 30 14 16 18...
Suppose we have the following data 12 17 13 25 16 21 30 14 16 18 to find the average value of 10% trimmed mean. Which number will be removed from calculation?
Part A. In the past it has been found that the arrival time have a population...
Part A. In the past it has been found that the arrival time have a population mean value of μ = 13 and a population standard deviation of σ = 6.26. Using the given data, test whether this mean has changed. Use the critical value approach to test the hypothesis. The significance level alpha is set at 0.05 . The original data of time taken is given below. Show the process by using excel (Formulas). Part B. Test the hypothesis...
A research council wants to estimate the mean length of time (in minutes) that the average...
A research council wants to estimate the mean length of time (in minutes) that the average U.S. adult spends watching television using digital video recorders (DVR’s) each day. To determine the estimate, the research council takes random samples of 35 U.S. adults and obtains the following times in minutes. 24 27 26 29 33 21 18 24 23 34 17 15 19 23 25 29 36 19 18 22 16 45 32 12 24 35 14 40 30 19 14...
A research council wants to estimate the mean length of time (in minutes) that the average...
A research council wants to estimate the mean length of time (in minutes) that the average U.S. adult spends watching television using digital video recorders (DVR’s) each day. To determine the estimate, the research council takes random samples of 35 U.S. adults and obtains the following times in minutes. 24 27 26 29 33 21 18 24 23 34 17 15 19 23 25 29 36 19 18 22 16 45 32 12 24 35 14 40 30 19 14...
A research council wants to estimate the mean length of time (in minutes) that the average...
A research council wants to estimate the mean length of time (in minutes) that the average U.S. adult spends watching television using digital video recorders (DVR’s) each day. To determine the estimate, the research council takes random samples of 35 U.S. adults and obtains the following times in minutes. 24 27 26 29 33 21 18 24 23 34 17 15 19 23 25 29 36 19 18 22 16 45 32 12 24 35 14 40 30 19 14...
You are interested in finding a 90% confidence interval for the average commute that non-residential students...
You are interested in finding a 90% confidence interval for the average commute that non-residential students have to their college. The data below show the number of commute miles for 14 randomly selected non-residential college students. Round answers to 3 decimal places where possible. 10 21 26 20 18 12 11 16 17 6 14 25 25 6 a. To compute the confidence interval use a ? z t  distribution. b. With 90% confidence the population mean commute for non-residential college...
You are interested in finding a 90% confidence interval for the average commute that non-residential students...
You are interested in finding a 90% confidence interval for the average commute that non-residential students have to their college. The data below show the number of commute miles for 14 randomly selected non-residential college students. Round answers to 3 decimal places where possible. 10 21 26 20 18 12 11 16 17 6 14 25 25 6 a. To compute the confidence interval use a ? z t  distribution. b. With 90% confidence the population mean commute for non-residential college...
A sample of 90 women is​ obtained, and their heights​ (in inches) and pulse rates​ (in...
A sample of 90 women is​ obtained, and their heights​ (in inches) and pulse rates​ (in beats per​ minute) are measured. The linear correlation coefficient is 0.285 and the equation of the regression line is ModifyingAbove y with = 17.6 + 0.930 x​, where x represents height. The mean of the 90 heights is 63.4 in and the mean of the 90 pulse rates is 75.7 beats per minute. Find the best predicted pulse rate of a woman who is...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT