Question

A publisher wants to estimate the mean length of time? (in minutes) all adults spend reading newspapers. To determine this? estimate, the publisher takes a random sample of 15 people and obtains the results below. From past? studies, the publisher assumes standard deviation is 1.7 minutes and that the population of times is normally distributed.

9

10

6

8

11

7

10

8

7

12

12

11

9

9

9

Construct the? 90% and? 99% confidence intervals for the population mean. Which interval is? wider? If? convenient, use technology to construct the confidence intervals.

The? 90% confidence interval is

Answer #1

From the given data, the following statistics are calculated:

n = sample size = 15

= sample mean = 138/15 = 9.2

Given:

= Population SD = 1.7

SE = /

= 1.7/ = 0.4389

(a)

= 0.10

From Table, critical values of Z = 1.645

90 % Confidence Interval:

Z SE

= 9.2 (1.645 X 0.4389)

= 9.2 0.7221

= (8.4779 , 9.9221)

The 90 % confidence interval is:

8.4779 < < 9.9221

(b) = 0.01

From Table, critical values of Z = 2.576

99 % Confidence Interval is:

9.2 (2.576 X 0.4389)

= 9.2 1.1306

= (8.0694 , 10.3306)

The 99 % Confidence Interval is:

8.0694 < < 10.3306

(c)

99 % Confidence Interval is wider.

A publisher wants to estimate the mean length of time (in
minutes) all adults spend reading newspapers. To determine this
estimate, the publisher takes a random sample of 15 people and
obtains the results below. From past studies, the publisher
assumes sigma σ is 2.1 minutes and that the population of times is
normally distributed. 6 9 6 11 6 7 10 12 8 6 8 9 10 10 9
Construct the 90% and 99% confidence intervals for the
population...

A publisher wants to estimate the mean length of time (in
minutes) all adults spend reading newspapers. To determine this
estimate, the publisher takes a random sample of 15 people and
obtains the results below. From past studies, the publisher
assumes sigma is 2.1 minutes and that the population of times is
normally distributed.
6 10 12 6 6 10 7 7 8 8 11 11 9 10 7
Construct the 90% and 99% confidence intervals for the
population mean....

A publisher wants to estimate the mean length of time (in
minutes) all adults spend reading newspapers. To determine this
estimate, the publisher takes a random sample of 15 people and
obtains the results below. From past studies, the publisher
assumes sigma is 2.7 minutes and that the population of times is
normally distributed. 12 12 6 9 11 12 9 9 11 10 6 6 12 6 6
Construct the 90% and 99% confidence intervals for the population
mean....

A publisher wants to estimate the mean length of time (in
minutes) all adults spend reading newspapers. To determine this
estimate, the publisher takes a random sample of 15 people and
obtains the results below. From past studies, the publisher
assumes sigma is 1.9 minutes and that the population of times is
normally distributed. 10 10 8 10 10 6 10 11 7 9 7 7 8 12 11
Construct the 90% and 99% confidence intervals for the population
mean....

publisher wants to estimate the mean length of time (in
minutes) all adults spend reading newspapers. To determine this
estimate, the publisher takes a random sample of 15 people and
obtains the results below. From past studies, the publisher
assumes sigmaσ is1.9 minutes and that the population of times is
normally distributed.
11
7
8
12
7
11
6
6
9
9
7
8
10
8
10
Construct the 90% and 99% confidence intervals for the
population mean. Which interval...

A publisher wants to estimate the mean length of time (in
minutes) all adults spend reading newspapers. To determine this
estimate, the publisher takes a random sample of 15 people and
obtains the results below. From past studies, the publisher
assumes
sigmaσ
is
2.22.2
minutes and that the population of times is normally
distributed.
99
88
99
1111
66
99
99
77
1212
77
1212
66
99
1212
99
Construct the 90% and 99% confidence intervals for the
population mean....

A research council wants to estimate the mean length of time (in
minutes) that the average U.S. adult spends watching television
using digital video recorders (DVR’s) each day. To determine the
estimate, the research council takes random samples of 35 U.S.
adults and obtains the following times in minutes.
24
27
26
29
33
21
18
24
23
34
17
15
19
23
25
29
36
19
18
22
16
45
32
12
24
35
14
40
30
19
14...

A research council wants to estimate the mean length of time (in
minutes) that the average U.S. adult spends watching television
using digital video recorders (DVR’s) each day. To determine the
estimate, the research council takes random samples of 35 U.S.
adults and obtains the following times in minutes.
24
27
26
29
33
21
18
24
23
34
17
15
19
23
25
29
36
19
18
22
16
45
32
12
24
35
14
40
30
19
14...

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. Which interval is? wider? If?
convenient, use technology to construct the confidence intervals. A
random sample of 39 gas grills has a mean price of ?$643.60 and a
standard deviation of ?$56.60 The? 90% confidence interval is __ ,
___

You want to estimate the mean time
college students spend watching online videos each day. The
estimate must be within 2 minutes of the population mean. Determine
the required sample size to construct a 99% confidence interval for
the population mean. Assume that the population standard deviation
is 4.4 minutes. Leave as an integer.

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 4 minutes ago

asked 5 minutes ago

asked 8 minutes ago

asked 10 minutes ago

asked 12 minutes ago

asked 17 minutes ago

asked 36 minutes ago

asked 40 minutes ago

asked 44 minutes ago

asked 48 minutes ago

asked 51 minutes ago

asked 53 minutes ago