Question

Suppose a randomly selected sample of n = 62 men has a mean foot length of...

Suppose a randomly selected sample of n = 62 men has a mean foot length of x = 28 cm, and the standard deviation of the sample is 3 cm. Calculate an approximate 95% confidence interval for the mean foot length of men. (Round the answers to one decimal place.)
to cm

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

Homework Answers

Answer #1

Solution :

Given that,

sample size = n = 62

Degrees of freedom = df = n - 1 = 62 - 1 = 61

t /2,df = 2.000

Margin of error = E = t/2,df * (s /n)

= 2.000 * ( 3/ 62)

= 0.8

The 95% confidence interval estimate of the population mean is,

- E < < + E

28 - 0.8 < < 28 + 0.8

27.2 < < 28.8

95% confidence interval for the mean: (27.2 , 28.8)

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
Imagine that 41 randomly selected Miramar students are asked to measure the length of one of...
Imagine that 41 randomly selected Miramar students are asked to measure the length of one of their feet. Suppose that the sample mean foot length (x) turns out to be 23.4 centimeters and that the sample standard deviation of the foot lengths turns out to be 5.1 centimeters. Find a 95% confidence interval for µ, the mean foot length among all Miramar students. Would you expect about 95% of all Miramar students to have foot lengths within this interval? Explain.
Assume that population mean is to be estimated from the sample described. Use the sample results...
Assume that population mean is to be estimated from the sample described. Use the sample results to approximate the margin of error and​ 95% confidence interval. Sample​ size, n=64​; sample​ mean, x overbare=83.0 ​cm; sample standard​ deviation, s=4.0 cm. The margin of error is ____ cm. ​(Round to one decimal place as​ needed.)
Suppose a landscaper wants to tell his clients the average length of the blades of grass...
Suppose a landscaper wants to tell his clients the average length of the blades of grass in a freshly mowed lawn. He measures the length of 50 randomly selected blades of grass immediately after mowing a lawn. He does not know the population standard deviation of grass lengths in the lawn, but his data are normally distributed with no outliers. The table contains a summary of his data. Sample size Sample mean (cm) Sample standard deviation (cm) ?n ?⎯⎯⎯x¯ ?s...
A simple random sample with n = 57 provided a sample mean of 22.5 and a...
A simple random sample with n = 57 provided a sample mean of 22.5 and a sample standard deviation of 4.4. (Round your answers to one decimal place.) (a) Develop a 90% confidence interval for the population mean.    to   (b) Develop a 95% confidence interval for the population mean.   to   (c) Develop a 99% confidence interval for the population mean.    to  
A sample of size = n 43 has sample mean = x 53.6 and sample standard...
A sample of size = n 43 has sample mean = x 53.6 and sample standard deviation = s 9.1 . Part: 0 / 2 0 of 2 Parts Complete Part 1 of 2 Construct a 90 % confidence interval for the population mean μ . Round the answers to one decimal place. A 90 % confidence interval for the population mean μ is
The state test scores for 12 randomly selected high school seniors are shown on the right....
The state test scores for 12 randomly selected high school seniors are shown on the right. Complete parts​ (a) through​ (c) below. Assume the population is normally distributed. 1425 1223 988 694 720 830 721 743 544 620 1442 943 ​(a)Find the sample mean. x=____ ​(Round to one decimal place as​ needed.) ​(b) Find the sample standard deviation. s=____ (Round to one decimal place as​ needed.) ​(c) Construct a 95% confidence interval for the population mean μ. A 95% confidence...
Thirty items are randomly selected from a population of 320 items. The sample mean is 29,...
Thirty items are randomly selected from a population of 320 items. The sample mean is 29, and the sample standard deviation 3. Develop a 80% confidence interval for the population mean. (Round the t-value to 3 decimal places. Round the final answers to 2 decimal places.)   The confidence interval is between and
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:...
Researchers are interested in the average length of men in a large population. They randomly select...
Researchers are interested in the average length of men in a large population. They randomly select 40 men from the population and measure their lengths to get average of 175 (cm). Scientists also know in advance that the lengths are approximately normal in standard deviation σ = 20 (cm). (a) Determine a 95% confidence interval for the average length of men for the entire population. (b) Determine the corresponding estimate at 99% confidence level.
In the study, n = 50 men were put on a diet. The men who dieted...
In the study, n = 50 men were put on a diet. The men who dieted lost an average of 7.2 kg, with a standard deviation of 3.5 kg. Question: Compute an approximate 95% confidence interval for the mean weight loss for the men who dieted. (Round your answer to the nearest hundredth.) An approximate 95% confidence interval is __to__kg
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT