Question
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?
Homework Answers
Answer #1
a)
df = n - 1 = 25 - 1 = 24
t critical value at 0.20 significance level = 1.318
80% confidence interval for 
is
- t * S /
sqrt(n) < < + t * S /
sqrt(n)
33.16 - 1.318 * 9.44 / sqrt(25) < < 33.16 + 1.318
* 9.44 / sqrt(25)
30.672 < < 35.648
80% CI is ( 30.672 , 35.648 )
b)
Margin of error = t * S / sqrt(n)
= 1.318 * 9.44 / sqrt(25)
= 2.488
Not the answer you're looking for?
Similar Questions
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 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 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
11.0years?
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...
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 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...
39 29 20 33 28 25 23 34 37 39 26 34 33 42 32 42...
39 29 20 33 28
25 23 34 37 39
26 34 33 42 32
42 45 24 32 41
48 36 16 13 40
A survey of 25 randomly selected customers found the ages shown
(in years). The mean is 32.44 years and the standard deviation is
8.97 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...
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 27.08 and the standard
deviation is 11.73 The owner wants to know if the mean age of all
customers is 32 years old.
11
16
28
43
49
41
33
17
35
36
15
24
15
11
35
25
34
40
41
32
22
13
34
11
16
c) What is the value of the test statistic?
d) What is the P-value...
Data from a survey of 50 randomly selected customers found a mean age of 38.7 years...
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?
(32.57, 44.83)
(34.78, 44.60)
(36.96, 40.44)
(38.70, 44.83)
ADVERTISEMENT
Need Online Homework Help?
Get Answers For Free
Most questions answered within 1 hours.
ADVERTISEMENT
Active Questions
asked 1 minute ago
asked 13 minutes ago
asked 25 minutes ago
asked 29 minutes ago
asked 32 minutes ago
asked 36 minutes ago
asked 1 hour ago
asked 1 hour ago
asked 1 hour ago
asked 1 hour ago
asked 1 hour ago
asked 1 hour ago