The amount of time adults spend watching television is closely monitored by firm

Question

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 watching? television" would be normally? distributed? If? not, what shape would you expect the variable to? have?

Answer is The variable? "weekly time spent watching? television" is likely skewed? right, not normally distributed.

(b) According to a certain? survey, adults spend 2.35 hours per day watching television on a weekday. Assume that the standard deviation for? "time spent watching television on a? weekday" is 1.93 hours. If a random sample of 40 adults is? obtained, describe the sampling distribution of x overbar, the mean amount of time spent watching television on a weekday.

Answer is x overbar is approximately normal with mu Subscript x overbar ?x equals=2.35 and sigma Subscript x overbar ?x equals=0.305160

(c) Determine the probability that a random sample of 40 adults results in a mean time watching television on a weekday of between 2 and 3 hours.

The probability is 0.8577

(d) One consequence of the popularity of the Internet is that it is thought to reduce television watching. Suppose that a random sample of 35 individuals who consider themselves to be avid Internet users results in a mean time of 1.93 hours watching television on a weekday. Determine the likelihood of obtaining a sample mean of 1.93 hours or less from a population whose mean is presumed to be 2.35 hours.

The likelihood is???

PLEASE ANSWER PART D ONLY. ALL OTHER PARTS ARE ANSWERED. PLEASE SHOW WORK USING A TI-84. THANKS.

Explanation / Answer

The likelihood is

P(xbar<1.93) = P((xbar-mean)/(s/vn) <(1.93-2.35)/(1.93/sqrt(35)))

=P(Z< -1.287437)

=0.099 (from standard normal table)

by ti-84

we have to use normalcdf function on TI-84
we need to find
P(Xbar < 1.93)

First: Lower boundary = -1000 (any negative large would work)
Second: Upper boundary = 1.93
Third: Average = 2.35
Fourth: Standard Deviation = 1.93/sqrt(35) = 0.32622954232

Access the normalcdf function on the calculator by pressing 2nd.

Then press VARS to access the DISTR menu

Now enter the 4 important numbers in order

Press Enter to get your answer.

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