Question
A sample of 2000 condominiums sold in Kelana Jaya gave the mean price of RM750,999. The...
- A sample of 2000 condominiums sold in Kelana Jaya gave the mean price of RM750,999. The population standard deviation is known to be RM160,850.
|
Homework Answers
Answer #1
Not the answer you're looking for?
Similar Questions
Estimation & Confidence Interval 1. A sample of 1500 homes sold recently in a state gave...
Estimation & Confidence Interval
1. A sample of 1500 homes sold recently in a state gave the mean
price of homes equal to $287,350. The population standard deviation
of the prices of homes in this state is $53,850.
a. Calculate the margin error and construct a 90% confidence
interval for the mean price of all homes in this state.
b. Calculate the margin error and construct a 95% confidence
interval for the mean price of all homes in this state....
In a random sample of five people, the mean driving distance to work was 24.9 miles...
In a random sample of five people, the mean driving distance to
work was 24.9 miles and the standard deviation was 4.3 miles.
Assuming the population is normally distributed and using the
t-distribution, a 99%confidence interval for the population mean
mu is left parenthesis 16.0 comma 33.8 right parenthesis (and the
margin of error is 8.9). Through research, it has been found that
the population standard deviation of driving distances to work is
3.3 Using the standard normal distribution with...
A random sample of 100 credit sales in a department store showed a sample mean of...
A random sample of 100 credit sales in a department store showed
a sample mean of $64. The population standard deviation is known to
be $16.
A) Provide a 99% confidence interval for the population
mean.
B) Provide a 95% confidence interval for the population
mean.
C) Describe the effect of a lower confidence level on the margin
of error.
In a random sample of eleven cell phones, the mean full retail price was 401.00 and...
In a random sample of eleven cell phones, the mean full retail
price was 401.00 and the standard deviation was 166.00. Assume the
population is normally distributed and use the t-distribution to
find the margin of error and construct a 90% confidence interval
for the population mean μ.
1-Identify the margin of error.
2-Construct a 90% confidence interval for the population
mean.
A simple random sample with n = 52 provided a sample mean of 28.5 and a...
A simple random sample with
n = 52
provided a sample mean of 28.5 and a sample standard deviation
of 4.4. (Round your answers to one decimal place.)
(a)
Develop a 90% confidence interval for the population mean.
to
(b)
Develop a 95% confidence interval for the population mean.
to
(c)
Develop a 99% confidence interval for the population mean.
to
(d)
What happens to the margin of error and the confidence interval
as the confidence level is increased?
As...
Let the following sample of 8 observations be drawn from a normal population with unknown mean...
Let the following sample of 8 observations be drawn from a
normal population with unknown mean and standard deviation: 16, 26,
20, 14, 23, 10, 12, 29. [You may find it useful to
reference the t table.]
a. Calculate the sample mean and the sample
standard deviation. (Round intermediate calculations to at
least 4 decimal places. Round "Sample mean" to 3 decimal places and
"Sample standard deviation" to 2 decimal places.)
b. Construct the 95% confidence interval for
the population...
In a random sample of 13 microwave ovens, the mean repair cost was $70.00 and the...
In a random sample of 13 microwave ovens, the mean repair cost
was $70.00 and the standard deviation was $15.40.
Using the standard normal distribution with the appropriate
calculations for a standard deviation that is known, assume the
population is normally distributed, find the margin of error and
construct a 98% confidence interval for the population mean. A
98% confidence interval using the t-distribution was
(58.5,81.5).
Find the margin of error of the population mean.
Find the confidence interval of...
1) Use the given confidence interval to find the margin of error and the sample mean....
1) Use the given confidence interval to find the margin of error
and the sample mean. (12.8,19.8)
2) In a random sample of four microwave ovens, the mean repair
cost was $60.00 and the standard deviation was $12.00. Assume the
population is normally distributed and use a t-distribution to
construct a 99% confidence interval for the population mean ?.
What is the margin of error of ? Interpret the results.
A simple random sample with n=50 provided a sample mean of 22.5 and a sample standard...
A simple random sample with n=50 provided a sample mean of 22.5
and a sample standard deviation of 4.2 .
a. Develop a 90% confidence interval for the
population mean (to 1 decimal).
( , )
b. Develop a 95% confidence interval for the
population mean (to 1 decimal).
( , )
c. Develop a 99% confidence interval for the
population mean (to 1 decimal).
( , )
d. What happens to the margin of error and the
confidence interval...
A random sample of 81 patients with end-stage renal disease gave a sample mean hemoglobin level...
A random sample of 81 patients with end-stage renal disease gave
a sample mean hemoglobin level of 10.5 g/dl with sample standard
deviation 2.79 g/dl. a) (5 pts) Calculate the Standard Error of the
sample mean b) (10 pts) Construct a 90% confidence interval for the
population mean hemoglobin level based on sample data. c) (5 pts)
Interpret the confidence interval you constructed in part (b) based
on the context of the problem. d) (5 pts) Now if the researchers...
ADVERTISEMENT
Need Online Homework Help?
Get Answers For Free
Most questions answered within 1 hours.
ADVERTISEMENT
Active Questions
asked 9 minutes ago
asked 11 minutes ago
asked 18 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 2 hours ago
asked 2 hours ago