Question
In: Statistics and Probability
A survey of 25 randomly sampled judges employed by the state of Florida found that they...
A survey of 25 randomly sampled judges employed by the state of Florida found that they earned an average wage (including benefits) of $67.00 per hour. The sample standard deviation was $5.75 per hour.
- What is the best estimate of the population mean?
- Develop a 95% confidence interval for the population mean wage (including benefits) for these employees. (Round your answers to 2 decimal places.)
- How large a sample is needed to assess the population mean with an allowable error of $1.00 at 98% confidence? (Round up your answer to the next whole number.)
Solutions
orchestra answered 1 year agoRelated Solutions
A survey of 25 randomly sampled judges employed by the state of Florida found that they...
A survey of 25 randomly sampled judges employed by the state of
Florida found that they earned an average wage (including benefits)
of $59.00 per hour. The sample standard deviation was $6.05 per
hour. (Use t Distribution Table.)
What is the best estimate of the population mean?
Develop a 99% confidence interval for the population mean wage
(including benefits) for these employees. (Round your
answers to 2 decimal places.)
How large a sample is needed to assess the population mean...
A survey of 20 randomly sampled judges employed by the state of Florida found that they...
A survey of 20
randomly sampled judges employed by the state of Florida found that
they earned an average wage (including benefits) of $63.00 per
hour. The sample standard deviation was $5.90 per hour. (Use
t Distribution Table.)
What is the best estimate of the population mean?
Develop a 98% confidence interval for the population mean wage
(including benefits) for these employees. (Round
your answers to 2 decimal places.)
Confidence interval
for the population mean wage is between and
How...
A survey of 24 randomly sampled judges employed by the state of Florida found that they...
A survey of 24
randomly sampled judges employed by the state of Florida found that
they earned an average wage (including benefits) of $57.00 per
hour. The sample standard deviation was $6.02 per hour. (Use
t Distribution Table.)
What is the best estimate of the population mean?
Develop a 98% confidence interval for the population mean wage
(including benefits) for these employees. (Round your
answers to 2 decimal places.)
A survey found that 34% of 875 randomly sampled teens said that their parents checked to...
A survey found that 34% of 875 randomly sampled teens said that
their parents checked to see what Web sites they visited. The
following year the same question posed to 800 teens found 42%
reporting such checks. Do these results provide evidence that more
parents are checking?
need: null and alternative hypothesis, z value, p value,
anything else that. might help
In a survey of post-operative infection, 25 surgical patient records are randomly sampled at the Minto...
In a survey of post-operative infection, 25 surgical patient
records are randomly sampled at the Minto Hospital. Of these, 21 of
the patients were found to have not suffered from a post-operative
infection. Nationally, 94% of surgeries do not result in the
patient suffering any post-operative infections.
(a) Briefly describe the population, random variable, and
parameter associated with this survey.
(b) Carry out a test to see if there is any evidence that the
Minto Hospital deviates from the national...
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?
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