1, A random sample of the closing stock prices in dollars for a company in a recent year is listed below. Assume that is $2.24. Construct the 90% and 99% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals.

16.39

19.57

17.99

16.05

16.28

22.28

20.27

22.78

17.16

19.18

19.64

17.05

21.03

17.03

18.46

17.18

The90% confidence interval is $,$.

(Round to two decimal places asneeded.)

2, A publisher wants to estimate the mean length of time(in minutes) all adults spend reading newspapers. To determine thisestimate, the publisher takes a random sample of 15 people and obtains the results below. From paststudies, the publisher assumes is 2.3 minutes and that the population of times is normally distributed.

10

12

6

6

9

12

12

9

8

11

11

7

12

7

8

3, Construct the90% and99% confidence intervals for the population mean. Which interval iswider? Ifconvenient, use technology to construct the confidence intervals.

The90% confidence interval is

(nothing, nothing).

(Round to one decimal place asneeded.)

4, You are given the sample mean and the population standard deviation. Use this information to construct the90% and95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. From a random sample of 74 dates, the mean record high daily temperature in a certain city has a mean of 84.59F. Assume the population standard deviation is 14.30F.

The90% confidence interval is

(nothing, nothing).

(Round to two decimal places asneeded.)

5, Use the confidence interval to find the estimated margin of error. Then find the sample mean.

A store manager reports a confidence interval of (46.8,79.6) when estimating the mean price(in dollars) for the population of textbooks.

The estimated margin of error is

nothing.

(Type" an integer or adecimal.)

6, Use the confidence interval to find the estimated margin of error. Then find the sample mean.

A biologist reports a confidence interval of (2.2,4.0) when estimating the mean height(in centimeters) of a sample of seedlings.

The estimated margin of error is

nothing.

7, Use the confidence interval to find the margin of error and the sample mean.

(0.232,0.370)

The margin of error is

nothing.

8, Construct the confidence interval for the population mean .

c=0.90,

x=4.2,

=0.4,

and

n=40

A 90% confidence interval for is

,.

(Round to two decimal places asneeded.)

9, According to onestudy, brain weights of men are normally distributed with a mean of

1.50 kg and a standard deviation of 0.13kg. Use the data to answer questions(a) through(e).

a. Determine the sampling distribution of the sample mean for samples of size 3. The mean of the sample means is

x=nothing.

10, For the provided sample,mean, samplesize, and population standarddeviation, complete parts(a) through(c) below. x=27, n=64, =3

a. Find a95% confidence interval for the population mean.

The95% confidence interval is from

nothing

to

nothing.

(Round to two decimal places asneeded.)

11, A random sample of 20 recent weddings in a country yielded a mean wedding cost of $26,334.75.

Assume that recent wedding costs in this country are normally distributed with a standard deviation of

$8200. Complete parts(a) through(c) below.

a. Determine a95% confidence interval for the meancost,

,

of all recent weddings in this country.

The95% confidence interval is from

$nothing

to

$nothing.

(Round to the nearest cent asneeded.)

12, Use the standard normal distribution or thet-distribution to construct a

95% confidence interval for the population mean. Justify your decision. If neither distribution can beused, explain why. Interpret the results. In a random sample of 19 mortgageinstitutions, the mean interest rate was 3.64% and the standard deviation was 0.39%. Assume the interest rates are normally distributed.

Which distribution should be used to construct the confidenceinterval?

A. Use at-distribution because the interest rates are normally distributed and is known.

B. Use at-distribution because it is a randomsample, isunknown, and the interest rates are normally distributed.

C. Use a normal distribution because the interest rates are normally distributed and

is known.

D. Use a normal distribution because n<30

and the interest rates are normally distributed.

E. Cannot use the standard normal distribution or thet-distribution because isunknown, n<30, and the interest rates are not normally distributed. Select the correct choice belowand, ifnecessary, fill in any answer boxes to complete your" choice.

A. The 95% confidence interval is

(nothing, nothing).

(Round to two decimal places asneeded.)

B. Neither distribution can be used to construct the confidence interval.

13, The gas mileage(in miles pergallon) of 27 randomly selected sports cars are listed in the accompanying table. Assume the mileages are not normally distributed. Use the standard normal distribution or thet-distribution to construct a 95% confidence interval for the population mean. Justify your decision. If neither distribution can beused, explain why. Interpret the results.

LOADING...

Click the icon to view the sports car gas mileage. Let

be the population standard deviation and let n be the sample size. Which distribution should be used to construct the confidenceinterval?

Neither distribution can neitherdistributioncan

be used to construct the confidenceinterval, since the population is not normally distributed and n less than 30. thepopulationisnotnormallydistributedandn<30.

Identify the confidence interval. Select the correct choice belowand, ifnecessary, fill in the answer box to complete your" choice.

A.

,

(Round to one decimal place asneeded.)

B. Neither the standard normal distribution nor thet-distribution can be used to construct the interval.

Use the standard normal distribution or thet-distribution to construct a 95% confidence interval for the population mean. Justify your decision. If neither distribution can beused, explain why. Interpret the results.

14, In a recentseason, the population standard deviation of the yards per carrying for all running backs was

1.33. The yards per carrying of 25 randomly selected running backs are shown below. Assume the yards per carrying are normally distributed.

2.1

4.7

3.9

6.9

5.2

6.4

2.8

2.7

4.8

1.9

2.3

7.4

3.6

6.3

4.5

6.1

6.2

4.1

6.8

3.4

3.4

7.3

3.7

4.7

3.8

Which distribution should be used to construct the confidenceinterval?

A. Use at-distribution because n<30 and is known.

B. Use a normal distribution because is known and the data are normally distributed.

Your answer is correct.

C. Use a normal distribution because n<30, the data are normally distributed and is unknown.

D. Use at-distribution because n<30 and is unknown.

E. Cannot use the standard normal distribution or thet-distribution because isunknown, n<30,

and the data are not normally distributed.

15, Select the correct choice belowand, ifnecessary, fill in any answer boxes to complete your" choice.

A. The 95% confidence interval is

(nothing, nothing).

(Round to two decimal places asneeded.)

B. Neither distribution can be used to construct the confidence interval.

Use the given confidence interval to find the margin of error and the sample proportion.

(0.632,0.660)

E=nothing

(Type" an integer or adecimal.)

16, In a survey of 2201 adults in a recentyear, 1285 say they have made a NewYear's resolution.

Construct90% and95% confidence intervals for the population proportion. Interpret the results and compare the widths of the confidence intervals.

The90% confidence interval for the population proportion p is

,.

(Round to three decimal places asneeded.)

17, In a survey of 2142 adults in a recentyear, 750 made a NewYear's resolution to eat healthier.

Construct90% and95% confidence intervals for the population proportion. Interpret the results and compare the widths of the confidence intervals.

The90% confidence interval for the population proportion p is

,.

(Round to three decimal places asneeded.)

18, In a survey of 3386 adults,1433 say they have started paying bills online in the last year.

Construct a99% confidence interval for the population proportion. Interpret the results.

A99% confidence interval for the population proportion is

,.

(Round to three decimal places asneeded.)

19, In a survey of 2311 adults, 748 say they believe in UFOs. Construct a 90% confidence interval for the population proportion of adults who believe in UFOs. A 90% confidence interval for the population proportion is

(nothing, nothing).

(Round to three decimal places asneeded.)