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The amount of time Americans spend watching television is closely monitored by firms such as \(\mathrm{AC}\) Nielsen because this helps determine advertising pricing for commercials. (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? (b) According to the American Time Use Survey, adult Americans 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 adult Americans is obtained, describe the sampling distribution of \(\bar{x},\) the mean amount of time spent watching television on a weekday. (c) Determine the probability that a random sample of 40 adult Americans results in a mean time watching television on a weekday of between 2 and 3 hours. (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 1.89 hours watching television on a weekday. Determine the likelihood of obtaining a sample mean of 1.89 hours or less from a population whose mean is presumed to be 2.35 hours. Based on the result obtained, do you think avid Internet users watch less television?

Step by step solution

01

Analyze the Distribution Shape

Consider whether the variable 'weekly time spent watching television' would be normally distributed. Typically, this kind of data is not normally distributed because viewing habits can vary greatly among individuals. Expect a right-skewed distribution since more people watch a moderate amount of television while fewer watch excessive amounts.

02

Define the Sampling Distribution

Use the central limit theorem to describe the sampling distribution of \(\bar{x}\), the mean amount of time spent watching television on a weekday. Given that the population mean (\(\mu\)) is 2.35 hours and the standard deviation (\(\sigma\)) is 1.93 hours, for a sample size \(n = 40\), the sampling distribution of \(\bar{x}\) will be approximately normal with mean \(\mu_{\bar{x}} = 2.35\) hours and standard deviation \(\sigma_{\bar{x}} = \frac{1.93}{\sqrt{40}} \).

03

Calculate Standard Error

Calculate the standard error of the mean: \[ \sigma_{\bar{x}} = \frac{1.93}{\sqrt{40}} = \frac{1.93}{6.32} \approx 0.305 \]

04

Determine Probability for Mean Between 2 and 3 Hours

Convert the values 2 and 3 hours to z-scores, then find the probability that the sample mean falls between these values. Use \( z = \frac{(x - \mu_{\bar{x}})}{\sigma_{\bar{x}}} \). For \( x = 2 \): \[ z = \frac{(2 - 2.35)}{0.305} \approx -1.15 \]. For \( x = 3 \): \[ z = \frac{(3 - 2.35)}{0.305} \approx 2.13 \]. Using z-tables, find the probabilities corresponding to these z-scores and subtract them: \[ P(-1.15 < z < 2.13) \approx 0.9830 - 0.1251 = 0.8579 \].

05

Calculate Likelihood of Mean 1.89 or Less

Convert 1.89 hours to a z-score and find the corresponding probability. \( z = \frac{(1.89 - 2.35)}{0.305} \approx -1.51 \). Using z-tables, find the probability: \[ P(z \leq -1.51) \approx 0.0655 \].

06

Conclusion on Internet Users' TV Watching

Since the probability (0.0655) of obtaining a sample mean of 1.89 hours or less is relatively low, it suggests that avid Internet users might indeed watch less television than the general population.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Central Limit Theorem

Imagine we have a lot of different groups of people, and we measure how much TV they watch in a week. If we take the average TV watching time of each group, the distribution of these averages will start to form a normal shape. This happens even if the original TV watching times aren't normal. This is called the Central Limit Theorem (CLT). Basically, the more groups you look at, the more the average TV watching time looks like a bell curve. This is especially true if the group size is big, like 30 or more. The CLT helps us predict what the overall average will be, even if the data starts out messy.

Z-Score

A z-score helps you understand how far a data point is from the mean, compared to the overall data spread. Think of it this way: if you know the average TV watching time is 2.35 hours, and someone watches 3 hours, is that unusual? Z-scores answer this. They take your data point, subtract the mean, and then divide by how spread out the data is (standard deviation). It tells you how many 'standard deviations' away your point is. If z is 0, it's right at the mean. A high z-score (like 2 or more) is unusual, meaning the person watches a lot more TV than average. A low z-score (like -2) means they watch a lot less.

Standard Error

Standard error measures how much the average (mean) is expected to vary from sample to sample. If you're checking different groups of people’s TV watching times, the averages of these groups won’t be exactly the same. Standard error tells us how much those group averages are likely to differ. It's calculated using the standard deviation, which measures how spread out individual watching times are, divided by the square root of the sample size. For example, if the standard deviation is 1.93 hours and the sample size is 40, the standard error tells us how close our sample mean is to the actual population mean.

Right-Skewed Distribution

When data is right-skewed, most people have lower values, but there are a few really high values that stretch the distribution to the right. Imagine measuring weekly TV watching time. Most people might watch between 1 to 4 hours a day. However, some might watch way more, say 8 hours a day. These few 'TV fanatics' cause the distribution to be right-skewed. The peak of the data will be towards the left (lower TV hours), with a long tail stretching to the right (higher TV hours). In our case, TV watching times are right-skewed, because while many people watch moderate amounts, some watch a lot, pulling the average up.