Question
In: Statistics and Probability
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 value t* be for a 95% confidence interval with 99 degrees of freedom?
e. What would the margin of error be for a 95% confidence interval for this data (with the sample size of 25)?
f. Construct a 95% confidence interval for the mean age of all customers for this data (with the sample size of 25) assuming that all conditions and assumptions have been met for the confidence interval.
g. How would the confidence interval have changed if the standard deviation of the 25 randomly selected customers was 15 instead of 9.84?
Solutions
orchestra answered 2 years agoRelated 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 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?...
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
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parts a through f.
A survey of 25 randomly selected psychiatrists found an average hourly wage (including benefits) of $65.00...
A survey of 25 randomly selected psychiatrists found an average
hourly wage (including benefits) of $65.00 per hour. The sample
standard deviation was $6.25 per hour. 1. What is the population
mean? What is the best estimate of the population mean? 2. Develop
a 99% confidence interval for the population mean wage (including
benefits) for these psychiatrists. 3. How large a sample is needed
to assess the population mean with an allowable error of $1.00 at
95% confidence?
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