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? O A. The variable "weekly time spent watching television" is likely normally distributed. OB. The variable "weekly time spent watching television" is likely skewed right, not normally distributed. OC. The variable "weekly time spent watching television" is likely symmetric, but not normally distributed. D. The variable "weekly time spent watching television" is likely uniform, not normally distributed. O E. The variable "weekly time spent watching television" is likely skewed left, 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, the mean amount of time spent watching television on a weekday. with s-ando- (Round to six decimal places as needed.) (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 XI hours. The probability is .(Round to four decimal places as needed.) (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 1.91 hours watching television on a weekday. Determine the likelihood of obtaining a sample mean of 1.91 hours or less from a population whose mean is presumed to be 2.35 hours. The likelihood is (Round to four decimal places as needed.) Interpret this probability. Select the correct choice below and fill in the answer box within your choice. (Round to the nearest integer as needed.) O A. If 1000 different random samples of size n = 35 individuals from a population whose mean is assumed to be 2.35 hours is obtained, we would expect a sample mean of exactly 1.91 in about of the samples. B. If 1000 different random samples of size n= 35 individuals from a population whose mean is assumed to be 235 hours is obtained, we would expect a sample mean of 1.91 or more in about of the samples. OC. If 1000 different random samples of size n= 35 individuals from a population whose mean is assumed to be 2.35 hours is obtained, we would expect a sample mean of 1.91 or less in about of the samples. Based on the result obtained, do you think avid Internet users watch less television? Yes No

Answer 1

(a)  The amount of time spent watching television is not normally distributed, because there will be a few people who watch television unusually long (outliers) which makes the distribution skewed to the right.Hence correct option is B

(b) According to a survey, adults spend 2.35 hours per day watching television on a weekday. The standard deviation is 1.93 hours. A random sample of 40 adults is obtained. The sampling distribution of the mean amount of time spent watching television on a weekday will be,

Central limit theorem: If the sample size is large (30 or more), then the sampling distribution of the sample mean T is approximately normal. The sample size of 40 is at least 30, thus the sampling distribution of the sample mean T is then approximately normal by the central limit theorem. The sampling distribution of the sample mean has mean y and standard deviation Hy = u= 2.35 1.93 0.305160 V40 0]=

Central limit theorem: If the sample size is large (30 or more), then the sampling distribution of the sample mean T is approximately normal. The sample size of 40 is at least 30, thus the sampling distribution of the sample mean T is then approximately normal by the central limit theorem. The sampling distribution of the sample mean has mean y and standard deviation Hy = u= 2.35 1.93 0.305160 V40 0]=

is approximately normal (n>30)

T

$\mu _{x}=2.35$

$\sigma _{x}=0.305160$

(c) The probability that a random sample of 40 adults results in a mean time between 2 and 3 hours will be calculated as under.

The sampling distribution of the sample mean where mean = $\mu$ and standard deviation = $\frac{\alpha }{\sqrt{n}}$

The z value is the sample mean decreased by population mean and devided by standard deviation.So we have T- 2 -2.35 -1.15 o/vn 1.93/40 T-M 3-2.35 2.13 o/n 1.93/V40 We shall determine the corresponding probability by using probability table in the appendix P(2 X <3) = P(-1.15<2<2.13) = P(Z < 2.13) - P(Z <-1.15) = 0.9834 – 0.1251 = 0.8583

So the probability is 0.8583

(d) Considering that n=35

The sampling distribution of the sample mean has mean y and standard deviation The z-value is the sample mean decreased by the population mean, divided by the standard deviation: T-M 1.91-2.35 -1.3487 g/n 1.93/ 35 Determine the corresponding probability using normal probability table in the appendix. P(X < 1.91) = P(Z <-1.3487) = 0.0887 The probability is not small (larger than 0.05), thus the event is not unusual and this then indicates that avid Internet users do not watch less television.

The likelihood is 0.0887

If 1000 different random samples of size n=35 individuals from a population whose mean is assumed to be 2.35 hours is obtained, we would expect a sample mean of 1.91 or less in about 88 of the samples.

Based on the result obtained, avid internet users do not watch less television.