Question
A simple random sample of electronic components will be selected to test for the mean lifetime...
A
simple random sample of electronic components will be selected to
test for the mean lifetime in hours. Assume that component
lifetimes are normally distributed with population standard
deviation of 31 hours. How many components must be sampled so that
a 99% confidence interval will have margin of error of 6 hours?
Write only an integer as your answer.
Homework Answers
Answer #1
|
Refer Standard normal table/Z-table or use excel function "=NORM.S.INV((1-0.005))" to find the Z-value.
|
Round the value to the nearest integer.
Not the answer you're looking for?
Similar Questions
A SIMPLE RANDOM SAMPLE OF KITCHEN TOASTERS IS TO BE TAKEN TO DETERMINE THE MEAN OPERATIONAL...
A
SIMPLE RANDOM SAMPLE OF KITCHEN TOASTERS IS TO BE TAKEN TO
DETERMINE THE MEAN OPERATIONAL LIFETIME IN HOURS. ASSUME THAT THE
LIFETIMES ARE NORMALLY DISTRIBUTED WITH POPULATION STANDARD
DEVIATION 22 HOURS.
FIND THE SAMPLE SIZE NEEDED SO THAT 90% CONFIDENCE INTERVAL
FOR THE MEAN LIFETIME WILL HAVE A MARGIN OF ERROR OF 8.
A simple random sample of kitchen toasters is to be taken to determine the mean operational...
A simple random sample of kitchen toasters is to be taken to
determine the mean operational lifetime in hours. Assume that the
lifetimes are normally distributed with population standard
deviation σ=30 hours. Find the sample size needed so that a 90%
confidence interval for the mean lifetime will have a margin of
error of 7.
The lifetimes of a certain electronic component are known to be normally distributed with a mean...
The lifetimes of a certain electronic component are known to be
normally distributed with a mean of 1,400 hours and a standard
deviation of 600 hours.
For a random sample of 25 components, find the probability that
the sample mean is less than 1300 hours.
A
0.2033
B
0.8213
C
0.1487
D
0.5013
A simple random sample of size n is drawn. The sample mean, x overbar, is found...
A simple random sample of size n is drawn. The sample mean, x
overbar, is found to be 17.6, and the sample standard deviation,
s, is found to be 4.1. LOADING... Click the icon to view the table
of areas under the t-distribution. (a) Construct a 95%
confidence interval about mu if the sample size, n, is 35. Lower
bound: nothing; Upper bound: nothing (Use ascending order. Round
to two decimal places as needed.) (b) Construct a 95% confidence
interval...
A simple random sample of size n is drawn. The sample mean, x overbar, is found...
A simple random sample of size n is drawn. The sample mean, x
overbar, is found to be 17.7, and the sample standard deviation,
s, is found to be 4.8.
Click the icon to view the table of areas under the
t-distribution.
(a) Construct a 95% confidence interval about mu if the
sample size, n, is 35. Lower bound: nothing; Upper bound:
nothing (Use ascending order. Round to two decimal places as
needed.)
(b) Construct a 95% confidence interval about...
A simple random sample of size n is drawn. The sample mean, x overbar, is found...
A simple random sample of size n is drawn. The sample mean, x
overbar, is found to be 19.5, and the sample standard deviation,
s, is found to be 4.6. (a) Construct a 95% confidence interval
about mu if the sample size, n, is 35. Lower bound: nothing;
Upper bound: nothing (Use ascending order. Round to two decimal
places as needed.) (b) Construct a 95% confidence interval about
mu if the sample size, n, is 61. Lower bound: nothing; Upper...
A simple random sample of size n is drawn. The sample mean, x overbar, is found...
A simple random sample of size n is drawn. The sample mean, x
overbar, is found to be 18.4, and the sample standard deviation,
s, is found to be 4.4. LOADING... Click the icon to view the table
of areas under the t-distribution. (a) Construct a 95%
confidence interval about mu if the sample size, n, is 35. Lower
bound: nothing; Upper bound: nothing (Use ascending order. Round
to two decimal places as needed.) (b) Construct a 95% confidence
interval...
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...
A random sample of 45 households was selected as part of a study on electricity usage.,...
A random sample of 45 households was selected as part of a study
on electricity usage., and the number of kilowatt-hours (kWh) was
recoded for each household. The average usage was found to be 375
kWh. The population standard deviation is given as 51 KWh Assuming
the usage is normally distributed, find the margin of error for a
99% confidence interval for the mean. [3 marks] 12.5 14.9 19.6
22.8
A modification has been made to the production line which produces a certain electronic component. Past...
A modification has been made to the production line which
produces a certain electronic component. Past evidence shows that
lifetimes of the components have a population standard deviation of
700 hours. It is believed that the modification will not affect the
population standard deviation, but may affect the mean lifetime.
The previous mean lifetime was 3250 hours. A random sample of 50
components made with the modification is taken and the sample mean
lifetime for these components is found to...
ADVERTISEMENT
Need Online Homework Help?
Get Answers For Free
Most questions answered within 1 hours.
ADVERTISEMENT
Active Questions
asked 7 minutes ago
asked 36 minutes ago
asked 39 minutes ago
asked 46 minutes ago
asked 54 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