Question

Data from a survey of 50 randomly selected customers found a mean age of 38.7 years...

  1. Data from a survey of 50 randomly selected customers found a mean age of 38.7 years and the standard deviation was 6.13 years. Given that t* = 2.01, calculate the 95% confidence interval for the mean?   
    1. (32.57, 44.83)
    2. (34.78, 44.60)
    3. (36.96, 40.44)
    4. (38.70, 44.83)
Math   Statistics-And-Probability   
0 0
Add a commentTranscribed image text
Next >

Homework Answers

Answer #1

Solution :

sample size = n = 50

Degrees of freedom = df = n - 1 = 49

t /2,df = 2.01

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

= 2.01 * (6.13 / 50)

Margin of error = E = 1.74

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

- E < < + E

38.7 - 1.74 < < 38.7 + 1.74

36.96 < < 40.44

c. (36.96, 40.44)

(,)

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 25 randomly selected customers found that their average age was 31.84 years with...
A survey of 25 randomly selected customers found that their average age was 31.84 years with a standard deviation of 9.84 years. What is the standard error of the mean? How would the standard error have changed if the sample size had been 100 instead of 25 (assuming that the mean and standard deviations were the same)? How many degrees of freedom would the t-statistic have for this set of data? What would the critical value t* be for a...
A survey of 25 randomly selected customers found the ages shown (in years). The mean is...
A survey of 25 randomly selected customers found the ages shown (in years). The mean is 33.00 years and the standard deviation is 8.74 years. a) Construct a 95% confidence interval for the mean age of all customers, assuming that the assumptions and conditions for the confidence interval have been met. b) How large is the margin of error? c) How would the confidence interval change if you had assumed that the standard deviation was known to be 9.0 years?
A survey of 25 randomly selected customers found the ages shown​ (in years). The mean is...
A survey of 25 randomly selected customers found the ages shown​ (in years). The mean is 31.04 years and the standard deviation is 9.00 years.​ a) Construct a 99​% confidence interval for the mean age of all​ customers, assuming that the assumptions and conditions for the confidence interval have been met. ​b) How large is the margin of​ error? ​c) How would the confidence interval change if you had assumed that the standard deviation was known to be 9.0 ​years?
A survey of 25 randomly selected customers found the ages shown​ (in years). The mean is...
A survey of 25 randomly selected customers found the ages shown​ (in years). The mean is 32.88 years and the standard deviation is 8.96 years. ​a) Construct a 90​% confidence interval for the mean age of all​ customers, assuming that the assumptions and conditions for the confidence interval have been met. ​b) How large is the margin of​ error? ​c) How would the confidence interval change if you had assumed that the standard deviation was known to be 9.0 ​years?
A survey of 25 randomly selected customers found the ages shown​ (in years). The mean is...
A survey of 25 randomly selected customers found the ages shown​ (in years). The mean is 32.32years and the standard deviation is 9.77years. ​a) Construct a 80​% confidence interval for the mean age of all​ customers, assuming that the assumptions and conditions for the confidence interval have been met. ​b) How large is the margin of​ error? ​c) How would the confidence interval change if you had assumed that the standard deviation was known to be 10.0 years?
A survey of 25 randomly selected customers found the ages shown​ (in years). The mean is...
A survey of 25 randomly selected customers found the ages shown​ (in years). The mean is 33.16 years and the standard deviation is 9.44 years. ​a) Construct a 80​% confidence interval for the mean age of all​ customers, assuming that the assumptions and conditions for the confidence interval have been met. ​b) How large is the margin of​ error? ​a) What is the confidence​ interval? ​b) What is the margin of​ error?
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...
A survey of 25 randomly selected customers found the ages shown​ (in years). The mean is...
A survey of 25 randomly selected customers found the ages shown​ (in years). The mean is 32.52 years and the standard deviation is 9.40 years. ​ 37 33 38 24 48 41 39 38 15 26 20 41 37 25 11 28 35 33 28 31 40 30 48 42 25 a) Construct a 99​% confidence interval for the mean age of all​ customers, assuming that the assumptions and conditions for the confidence interval have been met. (__,__) ​b) How...
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 survey of 25randomly selected customers found the ages shown​ (in years). The mean is 32.88years...
A survey of 25randomly selected customers found the ages shown​ (in years). The mean is 32.88years and the standard deviation is10.67 years .​a) Construct a80​% confidence interval for the mean age of all​ customers, assuming that the assumptions and conditions for the confidence interval have been met. ​b) How large is the margin of​ error? ​c) How would the confidence interval change if you had assumed that the standard deviation was known to be 11.0​years?
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT