8.1.29-TQuestion Help

The amount of time adults spend watching television is closely monitored by firms because this helps to determine advertising pricing for commercials. Complete parts(a) through(d).

(a) Do you think the variable"weekly time spent watchingtelevision" would be normallydistributed? Ifnot, what shape would you expect the variable tohave?

A.

The variable"weekly time spent watchingtelevision" is likely skewedleft, not normally distributed.

B.

The variable"weekly time spent watchingtelevision" is likelysymmetric, but not normally distributed.

C.

The variable"weekly time spent watchingtelevision" is likely skewedright, not normally distributed.

This is the correct answer.

D.

The variable"weekly time spent watchingtelevision" is likelyuniform, not normally distributed.

Your answer is not correct.

E.

The variable"weekly time spent watchingtelevision" is likely normally distributed.

(b) According to a certainsurvey, adults spend 2.45

2.45 hours per day watching television on a weekday. Assume that the standard deviation for"time spent watching television on aweekday" is 1.93

1.93 hours. If a random sample of 60

60 adults isobtained, describe the sampling distribution of x overbar

x, the mean amount of time spent watching television on a weekday.

x overbar

x is approximately normal

with mu Subscript x overbar

xequals

=

2.45

2.45 and sigma Subscript x overbar

xequals

=

0.249162

0.249162.

(Round to six decimal places asneeded.)

(c) Determine the probability that a random sample of 60

60 adults results in a mean time watching television on a weekday of between 2 and 3 hours.

The probability is

0.9509

0.9509. (Round to four decimal places asneeded.)

(d) One consequence of the popularity of the Internet is that it is thought to reduce television watching. Suppose that a random sample of 55

55 individuals who consider themselves to be avid Internet users results in a mean time of 2.10

2.10 hours watching television on a weekday. Determine the likelihood of obtaining a sample mean of 2.10

2.10 hours or less from a population whose mean is presumed to be 2.45

2.45 hours.

The likelihood is

0.0893

0.0893. (Round to four decimal places asneeded.)

Interpret this probability. Select the correct choice below and fill in the answer box within your choice.

(Round to the nearest integer asneeded.)

A.

If 1000 different random samples of size nequals

=55

55 individuals from a population whose mean is assumed to be 2.45

2.45 hours isobtained, we would expect a sample mean of 2.10

2.10 or more in about

nothing

of the samples.

B.

If 1000 different random samples of size nequals

=55

55 individuals from a population whose mean is assumed to be 2.45

2.45 hours isobtained, we would expect a sample mean of exactly 2.10

2.10 in about

nothing

of the samples.

C.

If 1000 different random samples of size nequals

=55

55 individuals from a population whose mean is assumed to be 2.45

2.45 hours isobtained, we would expect a sample mean of 2.10

2.10 or less in about

90

90 of the samples.

Your answer is not correct.

Based on the resultobtained, do you think avid Internet users watch lesstelevision?

Yes

Your answer is not correct.

No