Data on salaries in the public school system are published annually by a teachers' association. The mean annual salary of (public) classroom teachers is

$53.6

thousand. Assume a standard deviation of

$7.2

thousand. Complete parts (a) through (e) below.

Question content area bottom

Part 1

a. Determine the sampling distribution of the sample mean for samples of size 64.

The mean of the sample mean is

x=$enter your response here.

(Type an integer or a decimal. Do not round.)

Part 2

The standard deviation of the sample mean is

x=$enter your response here.

Part 3

b. Determine the sampling distribution of the sample mean for samples of size 256.

The mean of the sample mean is

x=$enter your response here.

Part 4

The standard deviation of the sample mean is

x=$enter your response here.

Part 5

c. Do you need to assume that classroom teacher salaries are normally distributed to answer parts (a) and (b)? Explain your answer.

A.

Yes, because

x

is only normally distributed if x is normally distributed.

B.

No, because if

x

is normally distributed, then x must be normally distributed.

C.

Yes, because the sample sizes are not sufficiently large so that

x

will be approximately normally distributed, regardless of the distribution of x.

D.

No, because the sample sizes are sufficiently large so that

x

will be approximately normally distributed, regardless of the distribution of x.

Part 6

d. What is the probability that the sampling error made in estimating the population mean salary of all classroom teachers by the mean salary of a sample of 64 classroom teachers will be at most $1000?

enter your response here

(Round to three decimal places as needed.)

Part 7

e. What is the probability that the sampling error made in estimating the population mean salary of all classroom teachers by the mean salary of a sample of 256 classroom teachers will be at most $1000?

enter your response here

(Round to three decimal places as needed.)

Suppose that the average time spent per day with digital media several years ago was

3

hours and

37

minutes. For last year, a random sample of

20

adults in a certain region spent the numbers of hours per day with digital media given in the accompanying table. Preliminary data analyses indicate that the t-interval procedure can reasonably be applied. Find and interpret a

99%

confidence interval for last year's mean time spent per day with digital media by adults of the region. (Note:

x=5.64

hr and

s=2.21

hr.)

Click here to view the digital media times.

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Click here to view page 1 of the t-table.

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Click here to view page 2 of the t-table.

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Question content area bottom

Part 1

The

99%

confidence interval is from

enter your response here

hour(s) to

enter your response here

hour(s).

(Round to two decimal places as needed.)

Part 2

Interpret the

99%

confidence interval. Select all that apply.

A.

99%

of all possible random samples of

20

adults in the region have mean amounts of time spent per day on digital media last year that are between the interval's bounds.

B.

With

99%

confidence, the mean amount of time spent per day on digital media last year by all adults in the region is between the interval's bounds.

C.

99%

of all adults in the region spent amounts of time per day on digital media last year that are between the interval's bounds.

D.

There is a

99%

chance that the mean amount of time spent per day on digital media last year by all adults in the region is between the interval's bounds.

a. Use the one-mean t-interval procedure with the sample mean, sample size, sample standard deviation, and confidence level given below to find a confidence interval for the mean of the population from which the sample was drawn.

b. Obtain the margin of error by taking half the length of the confidence interval.

c. Obtain the margin of error by using the formula

t/2sn.

x=32

n=25

s=2

confidence level=99%

Click here to view Page 1 of the table of t-values with area alpha to its right.

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Click here to view Page 2 of the table of t-values with area alpha to its right.

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Question content area bottom

Part 1

a. The

99%

confidence interval about

is

enter your response here

to

enter your response here.

(Round to three decimal places as needed.)

Part 2

b. The margin of error found by taking half the length of the confidence interval is

enter your response here.

(Round to three decimal places as needed.)

Part 3

c. The margin of error found by using the formula is

enter your response here.