Question
The king of a large castle would like to estimate the average age of knights currently...
The king of a large castle would like to estimate the average age of knights currently enrolled. Previous studies show a standard deviation of 2 years in knight ages.
a. If in a random sample of 25 knights at the castle, the average age x was found to be 23.2 years old, find a 95% confidence interval for the average age of all knights at the castle.
b. If the king wanted a 95% confidence interval for the average knight age that was no more than 1 year in width, how large a sample of knights would be needed?
Homework Answers
Answer #1
a)
95% Confidence Interval :-
X̅ ± Z( α /2) σ / √ ( n )
Z(α/2) = Z (0.05 /2) = 1.96
23.2 ± Z (0.05/2 ) * 2/√(25)
Lower Limit = 23.2 - Z(0.05/2) 2/√(25)
Lower Limit = 22.416
Upper Limit = 23.2 + Z(0.05/2) 2/√(25)
Upper Limit = 23.984
95% Confidence interval is ( 22.416 , 23.984
)
b)
Margin of error E = Width of confidence interval / 2 = 1 / 2 = 0.5
Sample size = (Z
/2
* 
/ E)2
= ( 1.96 * 2 / 0.5)2
= 61.5
Sample size = 62 (Rounded up to nearest integer)
Not the answer you're looking for?
Similar Questions
A college admissions director wishes to estimate the mean age of all students currently enrolled. In...
A college admissions director wishes to estimate the mean age of
all students currently enrolled. In a random sample of 22 students,
the mean age is found to be 21.4 years. From past studies, the ages
of enrolled students are normally distributed with a standard
deviation of 10.2 years. Construct a 90% confidence interval for
the mean age of all students currently enrolled.
1. What is the Critical value?
A college admissions director wishes to estimate the mean age of all students currently enrolled. In...
A college admissions director wishes to estimate the mean age of
all students currently enrolled. In a random sample of 19 students,
the mean age is found to be 23.7 years. From past studies, the ages
of enrolled students are normally distributed with a standard
deviation of 10.4 years. Construct a 90% confidence interval for
the mean age of all students currently enrolled.
1. The critical value:
2. The standard deviation of the sample mean:
3. The margin of error...
The makers of a soft drink want to identify the average age of its consumers. A...
The makers of a soft drink want to identify the average age of
its consumers. A sample of 16 consumers is taken. The average age
in the sample was 22.5 years with a standard
deviation of 5 years. Assume the population of consumer ages is
normally distributed.
a.Construct a 95% confidence interval for the average age of all
the consumers.
b.Construct an 80% confidence interval for the average age of
all the consumers.
c.Discuss why the 95% and 80% confidence...
PART I The owner of a local golf course wanted to determine the average age (in...
PART I
The owner of a local golf course wanted to determine the average
age (in years) of the golfers that played on the course. In a
random sample of 27 golfers that visited his course, the sample
mean was 47 years old and the standard deviation was 5.11 years.
Using this information, the owner calculated the confidence
interval of (45.3, 48.7) with a confidence level of 90% for the
average age. Which of the following is an appropriate
interpretation...
The proprieter of a boutique in New York wanted to determine the average age of his...
The proprieter of a boutique in New York wanted to determine the
average age of his customers. A random sample of 25 customers
revealed an average of 28 years with a standard deviation of 10
years. Determine a 95% confidence interval estimate for the average
age of all his customers.
In examining the credit accounts of a department store, an auditor would like to estimate the...
In examining the credit accounts of a department store, an
auditor would like to estimate the true mean account error (book
value – audited value). To do so, the auditor selected a random
sample of 50 accounts and found the mean account error to be $60
with a standard deviation of $30.
a. Construct a 95% confidence interval for the population mean
account error. What conclusion can be made from this confidence
interval?
b. How large a sample is actually...
A college admissions director wishes to estimate the mean age of all students currently enrolled. The...
A college admissions director wishes to estimate the mean age of
all students currently enrolled. The age of a random sample of 23
students is given below. Assume the ages are approximately normally
distributed. Use Excel to construct a 90% confidence interval for
the population mean age. Round your answers to two decimal places
and use increasing order.
Age
25.8
22.2
22.5
22.8
24.6
24.0
22.6
23.6
22.8
23.1
21.5
21.4
22.5
24.5
21.5
22.5
20.5
23.0
25.1
25.2
23.8...
The operations manager of a large production plant would like to estimate the average amount of...
The operations manager of a large production plant would like to
estimate the average amount of time workers take to assemble a new
electronic component. After observing a number of workers
assembling similar devices, she guesses that the standard deviation
is 6 minutes. How large a sample of workers should she take if she
wishes to estimate the mean assembly time to within 30 seconds with
a confidence level of 95%. (Be sure to use plenty of deicmal places
throughout...
For a certain group of cancer patients we are interested in estimating the mean age of...
For a certain group of cancer patients we are interested in
estimating the mean age of diagnosis. We would like a 95%
confidence level.
Assume the population standard deviation is 12 years.
a. What sample size do you recommend for an interval width of 5
years?
b. What sample size do you recommend for an interval width of 2
years?
We would like to estimate the average age in the population. Assume that standard deviation is...
We would like to estimate the average age in the population.
Assume that standard deviation is equal to 10 years. Compute the
optimal sample size in case of maximum tolerable error of 2 units
and significance level 0,05. Population size is not known.
ADVERTISEMENT
Need Online Homework Help?
Get Answers For Free
Most questions answered within 1 hours.
ADVERTISEMENT
Active Questions
asked 38 minutes ago
asked 56 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
asked 1 hour ago
asked 1 hour ago
asked 1 hour ago
asked 1 hour ago