Question

You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals. A random sample of 45 45 home theater systems has a mean price of $ 137.00 137.00. Assume the population standard deviation is $ 16.60 16.60. Construct a 90% confidence interval for the population mean. The 90% confidence interval is ( nothing , nothing ). (Round to two decimal places as needed.) Construct a 95% confidence interval for the population mean. The 95% confidence interval is ( nothing , nothing ). (Round to two decimal places as needed.) Interpret the results. Choose the correct answer below. A. With 90% confidence, it can be said that the population mean price lies in the first interval. With 95% confidence, it can be said that the population mean price lies in the second interval. The 95% confidence interval is wider than the 90%. B. With 90% confidence, it can be said that the population mean price lies in the first interval. With 95% confidence, it can be said that the population mean price lies in the second interval. The 95% confidence interval is narrower than the 90%. C. With 90% confidence, it can be said that the sample mean price lies in the first interval. With 95% confidence, it can be said that the sample mean price lies in the second interval. The 95% confidence interval is wider than the 90%.

Answer #1

Solution :

Given that,

Point estimate = sample mean =
= 137.00

Population standard deviation =
= 16.60

Sample size = n = 45

1) At 90% confidence level

= 1 - 90%

= 1 - 0.90 =0.10

/2
= 0.05

Z/2
= Z_{0.05 = 1.645}

Margin of error = E = Z/2
* (
/n)

= 1.645 * ( 16.60 /
45)

= 4.07

At 90% confidence interval estimate of the population mean is,

± E

137.00 ± 4.07

( 132.93, 141.07 )

2) At 95% confidence level

= 1 - 95%

= 1 - 0.95 =0.05

/2
= 0.025

Z/2
= Z_{0.025 = 1.96}

Margin of error = E = Z/2
* (
/n)

= 1.96 * ( 16.60 /
45)

= 4.85

At 95% confidence interval estimate of the population mean is,

± E

137.00 ± 4.85

( 132.15, 141.85)

correct option is =A

With 90% confidence, it can be said that the population mean price lies in the first interval. With 95% confidence, it can be said that the population mean price lies in the second interval. The 95% confidence interval is wider than the 90%

You are given the sample mean and the population standard
deviation. Use this information to construct the 90% and 95%
confidence intervals for the population mean. Interpret the results
and compare the widths of the confidence intervals. If convenient,
use technology to construct the confidence intervals.
A random sample of 35 home theater systems has a mean price of
$128.00 Assume the population standard deviation is $19.30
Construct a 90% confidence interval for the population
mean.
The 90% confidence interval...

You are given the sample mean and the population standard
deviation. Use this information to construct the 90% and 95%
confidence intervals for the population mean. Interpret the results
and compare the widths of the confidence intervals. If convenient,
use technology to construct the confidence intervals. A random
sample of 55 home theater systems has a mean price of $115.00.
Assume the population standard deviation is $17.70.
1. The 90% confidence interval is?
2. Interpret the results. Choose the correct...

You are given the sample mean and the sample standard deviation.
Use this information to construct the 90% and 95% confidence
intervals for the population mean. Interpret the results and
compare the widths of the confidence intervals. If convenient, use
technology to construct the confidence intervals. A random sample
of 5050 home theater systems has a mean price of $145.00145.00 and
a standard deviation is $18.7018.70.
The 90% confidence interval is
The 95% confidence interval is
Interpret the results. Choose...

You are given the sample mean and the population standard
deviation. Use this information to construct the 90% and 95%
confidence intervals for the population mean. Interpret the results
and compare the widths of the confidence intervals. From a random
sample of 33 business days, the mean closing price of a certain
stock was $111.13. Assume the population standard deviation is
$10.41. The 90% confidence interval is ( nothing, nothing).
(Round to two decimal places as needed.) The 95% confidence...

You are given the sample mean and the population standard
deviation. Use this information to construct the 90% and 95%
confidence intervals for the population mean. Interpret the results
and compare the widths of the confidence intervals.
From a random sample of
78
dates, the mean record high daily temperature in a certain city
has a mean of
85.43 degrees°F.
Assume the population standard deviation is
13.72 degrees°F.
The 90% confidence interval is
(nothing,nothing).
(Round to two decimal places as...

You are given the sample mean and the population standard
deviation. Use this information to construct the 90% and 95%
confidence intervals for the population mean. Interpret the results
and compare the widths of the confidence intervals. If convenient,
use technology to construct the confidence intervals.
A random sample of 60 home theater systems has a mean price of
$147.00. Assume the population standard deviation is $19.40.
Construct a 90% confidence interval for the population mean.
The 90% confidence interval...

You are given the sample mean and the population standard
deviation. Use this information to construct the90% and 95%
confidence intervals for the population mean. Interpret the results
and compare the widths of the confidence intervals. If convenient,
use technology to construct the confidence intervals. A random
sample of 6060 home theater systems has a mean price of
$129.00129.00. Assume the population standard deviation is
$15.9015.90.
Construct a 90% confidence interval for the population
mean.
The 90% confidence interval is...

You are given the sample mean and the population standard
deviation. Use this information to construct the 90% and 95%
confidence intervals for the population mean. Interpret the results
and compare the widths of the confidence intervals. From a random
sample of 72 dates, the mean record high daily temperature in a
certain city has a mean of 82.800F. Assume the
population standard deviation is 15.250F.
The 90% confidence interval is ( , ). (Round to two decimal
places as...

You are given the sample mean and the population standard
deviation. Use this information to construct the 90% and 95%
confidence intervals for the population mean. Interpret the results
and compare the widths of the confidence intervals. If convenient,
use technology to construct the confidence intervals.
A random sample of 40 home theater systems has a mean price of
$140.00 Assume the population standard deviation is $19.50.
The 90% confidence interval is?
The 95% confidence interval is?

You are given the sample mean and the population standard
deviation. Use this information to construct the 90% and 95%
confidence intervals for the population mean. Interpret the results
and compare the widths of the confidence intervals. If convenient,
use technology to construct the confidence intervals. A random
sample of 50 home theater systems has a mean price of $142.00.
Assume the population standard deviation is $18.70.
Construct a 90% confidence interval for the population
mean.

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 8 minutes ago

asked 47 minutes ago

asked 49 minutes ago

asked 57 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