Question

the amount of time americans spend watching television is closely monitored by firms such as a.c. nielsen because this helps to determine the advertising pricing for commercials. according to a recent survey, adult americans spend an average of 2.3 hours per day watching television on a weekday. assume the standard deviation for \time spent watching television on a weekday\ is 1.89 hours.
a) if a random sample of 50 adult americans is obtained, what is the probability that the sample mean time watching television is between 2 and 3 hours?
b) one consequence of the popularity of the internet is that it is thought to reduce television watching. suppose that a random sample of 40 individuals who consider themselves to be avid internet users results in a mean time of 1.74 hours watching television on a weekday. determine the likelihood of obtaining a sample mean of 1.74 hours or less from a population whose mean is presumed to be 2.3 hours.

Explanation:

Part B Solution:

Step 1: Identify the distribution parameters

We know that for the sampling distribution of the sample mean, the mean \(\mu_{\bar{x}}=\mu = 2.3\) hours (population mean), the standard deviation of the sample mean (standard error) is \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\). First, we need to find \(\sigma\). From part A, we can assume the population standard deviation \(\sigma\) (since the standard deviation for the time spent watching television on a weekday is given as 1.89 hours? Wait, no, wait. Wait, in part A, the problem was about a sample of 50, but here part B is a sample of 40. Wait, let's re - read part B: "Suppose that a random sample of 40 individuals who consider themselves to be avid Internet users results in a mean time of 1.74 hours watching television on a weekday. Determine the likelihood of obtaining a sample mean of 1.74 hours or less from a population whose mean is presumed to be 2.3 hours."

We assume that the population standard deviation \(\sigma\) is the same as the standard deviation for the time spent watching television on a weekday, which is \(\sigma = 1.89\) hours (from the problem statement: "the standard deviation for 'time spent watching television on a weekday' is 1.89 hours"). The sample size \(n = 40\), the population mean \(\mu=2.3\), and the sample mean \(\bar{x} = 1.74\).

The formula for the z - score of the sample mean is \(z=\frac{\bar{x}-\mu_{\bar{x}}}{\sigma_{\bar{x}}}\), where \(\mu_{\bar{x}}=\mu\) and \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\)

Step 2: Calculate the standard error

First, calculate the standard error \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}=\frac{1.89}{\sqrt{40}}\)

\(\sqrt{40}\approx6.3246\), so \(\sigma_{\bar{x}}=\frac{1.89}{6.3246}\approx0.2988\)

Step 3: Calculate the z - score

Now, calculate the z - score for \(\bar{x} = 1.74\):

\(z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}=\frac{1.74 - 2.3}{0.2988}=\frac{- 0.56}{0.2988}\approx - 1.87\)

Step 4: Find the probability using the z - table

We want to find \(P(\bar{X}\leq1.74)=P(Z\leq - 1.87)\)

Looking up the z - score of \(- 1.87\) in the standard normal z - table. The area to the left of \(z=-1.87\) is the probability we want.

From the z - table, the area to the left of \(z = - 1.87\) is approximately 0.0307.

Step 1: Identify the distribution parameters

We have a sample of \(n = 50\) adult Americans. The population mean \(\mu=2.3\) hours, the population standard deviation \(\sigma = 1.89\) hours. We want to find \(P(2\leq\bar{X}\leq3)\), where \(\bar{X}\) is the sample mean.

The mean of the sampling distribution of the sample mean \(\mu_{\bar{x}}=\mu = 2.3\), and the standard error \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}=\frac{1.89}{\sqrt{50}}\)

Step 2: Calculate the standard error

\(\sqrt{50}\approx7.0711\), so \(\sigma_{\bar{x}}=\frac{1.89}{7.0711}\approx0.2673\)

Step 3: Calculate the z - scores for 2 and 3

For \(x = 2\):
\(z_1=\frac{2 - 2.3}{0.2673}=\frac{- 0.3}{0.2673}\approx - 1.12\)

For \(x = 3\):
\(z_2=\frac{3 - 2.3}{0.2673}=\frac{0.7}{0.2673}\approx2.62\)

Step 4: Find the probabilities corresponding to the z - scores

We know that \(P(2\leq\bar{X}\leq3)=P(- 1.12\leq Z\leq2.62)\)

\(P(-1.12\leq Z\leq2.62)=P(Z\leq2.62)-P(Z\leq - 1.12)\)

From the z - table, \(P(Z\leq2.62)\approx0.9956\) and \(P(Z\leq - 1.12)\approx0.1314\)

So \(P(-1.12\leq Z\leq2.62)=0.9956 - 0.1314 = 0.8642\) (close to the given 0.8648, the difference is due to more precise z - table values or calculation)

Answer:

The likelihood (probability) is approximately \(0.0307\)

Part A (if we need to re - solve it for completeness, but the user's main issue was part B):