Question

In: Statistics and Probability

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 9.33
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 10.0 ​years?

2727

3939

2626

3838

2424

4343

2121

1414

3737

4949

1313

3939

3434

3939

2626

4141

3030

2828

2828

4646

3232

4040

4343

3535

3737

​a) What is the confidence​ interval?

​(nothing​,nothing​)

​b) What is the margin of​ error?

The margin of error is (Nothing)

​c) What is the confidence interval using the given population standard​ deviation? Select the correct choice below and fill in the answer boxes within your choice.

​(Round to two decimal places as​ needed.)

A.The new confidence interval ​(nothing​,nothing​)is narrower than the interval from part a.

B.The new confidence interval ​(nothing​,nothing​) is wider than the interval from part a.

Solutions

Expert Solution

a)

sample mean 'x̄= 33.160
sample size    n= 25
std deviation s= 9.330
std error ='sx=s/√n=9.33/√25= 1.8660
for 95% CI; and 24 df, value of t= 2.0639
margin of error E=t*std error    = 3.8512
lower bound=sample mean-E = 29.309
Upper bound=sample mean+E = 37.011
from above 95% confidence interval for population mean =(29.31 <μ<37.01

b)

margin of error E=t*std error    = 3.85

(please try 3.851 if required to 3 decimals)

c)

sample mean 'x̄= 33.160
sample size    n= 25
std deviation σ= 10.000
for 95 % CI value of z= 1.960
margin of error E=z*std error = 3.920
lower bound=sample mean-E= 29.24
Upper bound=sample mean+E= 37.08
from above 95% confidence interval for population mean =(29.24,37.08)

B.The new confidence interval ​ (29.24 , 37.08) is wider than the interval from part a.


Related Solutions

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 32.64 years and the standard deviation is 9.97 years. ​a) What is the standard error of the​ mean? ​b) 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.)
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 32.24 years and the standard deviation is 9.55 years.​ 30 41 48 36 20 36 37 36 26 27 17 23 39 35 21 10 31 42 32 41 43 28 47 25 35 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...
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 31.52 years and the standard deviation is 10.24 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 11.0 ​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 32.04 years and the standard deviation is 9.87 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? ​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 31.96 years and the standard deviation is 10.12 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 11.0 ​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 32.44 years and the standard deviation is 10.17 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 11.0 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 32.32 years and the standard deviation is 10.66 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 11.0 ​years?
A store owner surveyed 25 randomly selected customers and found the ages shown​ (in years). The...
A store owner surveyed 25 randomly selected customers and found the ages shown​ (in years). The mean is 33.24 and the standard deviation is 10.87. The owner wants to know if the mean age of all customers isis 29 years old. Use the given information to complete parts a through f.
2) A survey of 25 randomly selected customers found that their average age was 31.84 years...
2) A survey of 25 randomly selected customers found that their average age was 31.84 years with a standard deviation of 9.84 years. a. What is the standard error of the mean? b. 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)? c. How many degrees of freedom would the t-statistic have for this set of data? d. What would the critical...
A survey of 1060 randomly selected US teens ages 13 to 17 found that 605 of...
A survey of 1060 randomly selected US teens ages 13 to 17 found that 605 of them say they have made a new friend online. Find and interpret a 90% confidence interval for the proportion of US teens who have made a new friend online. Fill in the blanks a) The scenario above represents (write either test or CI): b) The population parameter (write either single proportion or single mean): c) The sample size: d) The sample statistic (round decimal...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT