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Do our children spend enough time enjoying the outdoors and playing with friends, or are they spending more time glued to the television, computer, and their cell phones? A random sample of 250 youth between the ages of 8 and 18 showed that 170 of them had a TV in their bedroom and that 120 had a video game console in their bedroom. a. Estimate the proportion of all 8 - to 18 -year-olds who have a TV in their bedroom, and calculate the margin of error for your estimate. b. Estimate the proportion of all 8 - to 18 -year-olds who have a video game console in their bedroom, and calculate the margin of error for your estimate.

Short Answer

Expert verified
Answer: The estimated proportions of 8- to 18-year-olds with a TV and a video game console in their bedrooms are 0.68 and 0.48, respectively, and the margin of error for both estimates is 0.056.

Step by step solution

01

a. Estimate the proportion of all 8- to 18-year-olds who have a TV in their bedroom.

First, calculate the proportion by dividing the number of youths with a TV by the total sample size: p(TV) = 170 / 250 p(TV) ≈ 0.68 (rounded to 2 decimal places) The estimated proportion of 8- to 18-year-olds who have a TV in their bedroom is 0.68.
02

Calculate the margin of error for the TV estimate.

Now, let's calculate the margin of error using the formula: Margin of Error = z * sqrt(p(1-p)/n) For TV, we have: z = 1.96 p = 0.68 n = 250 Margin of Error(TV) = 1.96 * sqrt(0.68 * (1 - 0.68) / 250) Margin of Error(TV) ≈ 0.056 (rounded to 3 decimal places) The margin of error for the estimate of the proportion of 8- to 18-year-olds who have a TV in their bedroom is 0.056.
03

b. Estimate the proportion of all 8- to 18-year-olds who have a video game console in their bedroom.

Again, calculate the proportion by dividing the number of youths with a video game console by the total sample size: p(Console) = 120 / 250 p(Console) ≈ 0.48 (rounded to 2 decimal places) The estimated proportion of 8- to 18-year-olds who have a video game console in their bedroom is 0.48.
04

Calculate the margin of error for the video game console estimate.

Using the same formula as before: Margin of Error = z * sqrt(p(1-p)/n) For video game console, we have: z = 1.96 p = 0.48 n = 250 Margin of Error(Console) = 1.96 * sqrt(0.48 * (1 - 0.48) / 250) Margin of Error(Console) ≈ 0.056 (rounded to 3 decimal places) The margin of error for the estimate of the proportion of 8- to 18-year-olds who have a video game console in their bedroom is 0.056. In conclusion, the estimated proportions of 8- to 18-year-olds with a TV and a video game console in their bedrooms are 0.68 and 0.48, respectively, and the margin of error for both estimates is 0.056.

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Key Concepts

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

Margin of Error
In statistics, the margin of error is a valuable measure that indicates the range within which we can expect the true population parameter to fall with a certain level of confidence. It's particularly important when we're drawing conclusions from a sample of a larger population, as in the given exercise with children's bedroom items like TVs and video game consoles.

The margin of error helps us understand the precision of our sample estimate. For example, if we find that 68% of a sample have a TV in their bedroom with a margin of error of 0.056, we can say with a certain level of confidence that the true proportion in the entire population is between 61.4% and 74.6%. A smaller margin of error indicates a more precise estimate, and it’s influenced by the sample size and the variability in the data.
Confidence Interval
Closely related to the margin of error is the confidence interval, which provides a range that is likely to contain the population parameter we are estimating. It's calculated by taking the sample statistic and adding or subtracting the margin of error. In the context of our exercise, if we have an estimated proportion (let's say for the television presence) of 0.68 and a margin of error of 0.056, the confidence interval would typically be computed as 0.68 ± 0.056.

However, even with this interval, we need to specify our level of confidence. Common confidence levels are 90%, 95%, and 99%, which correspond to z-scores used in calculating the margin of error. The confidence interval gives stakeholders a practical sense of where the true population value lies and the reliability of the data collected.
Sample Statistics
When we talk about sample statistics, we're looking at measures or characteristics calculated from the data obtained from a subset of the larger population. Examples of sample statistics include sample means, proportions, standard deviations, and others. In our exercise, the sample statistics would be the proportions of youths with a TV or a video game console in their bedroom (0.68 and 0.48, respectively).

It's crucial to understand that these statistics serve as estimates for the true population parameters, but they would vary depending on the sample taken. Therefore, it's important to use methods like the margin of error and confidence interval to express the uncertainty associated with these estimates.
Probability and Statistics
The fields of probability and statistics are fundamental to understanding data analysis, as they provide methods for making predictions about a population based on sample observations. Probability deals with the likelihood of events occurring, whereas statistics involves collecting, analyzing, interpreting, presenting, and organizing data.

In our exercise, we use probability to determine the likelihood that our sample statistic is a good estimate of the population parameter. Using statistical methods, we then calculate the margin of error and the confidence interval to express the reliability of our estimates with a specified level of assurance. These concepts are interconnected and essential for sound decision-making in various fields.

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Most popular questions from this chapter

A random sample of \(n\) measurements has been selected from a population with unknown mean \(\mu\) and known standard deviation \(\sigma=10 .\) Calculate the width of a \(95 \%\) confidence interval for \mu for the sample sizes given. What effect do the changing sample sizes have on the width of the interval? \(n=200\)

As Americans become more conscious of the importance of good nutrition, some researchers believe that we may be eating less red meat than we used to eat. To test this theory, a researcher selects two groups of 400 subjects each and collects the following sample information on the annual beef consumption now and 10 years ago: $$\begin{array}{lcc}\hline & \text { Ten Years Ago } & \text { This Year } \\\\\hline \text { Sample Mean } & 73 & 63 \\\\\text { Sample Standard Deviation } & 25 & 28 \\\\\hline\end{array}$$ a. The researcher would like to show that per-capita beef consumption has decreased in the last 10 years, so she needs to show that the difference in the averages is greater than 0 . Find a \(99 \%\) lower confidence bound for the difference in the average per-capita beef consumptions for the two groups. b. What conclusions can the researcher draw using the confidence bound from part a?

Suppose you want to estimate one of four parameters- \(\mu, \mu_{1}-\mu_{2}, p,\) or \(p_{1}-p_{2}-\) to within a given bound with a certain amount of confidence. Use the information given to find the appropriate sample size(s). Estimating \(p\) to within .04 with probability .95. You suspect that \(p\) is equal to some value between .1 and \(.3 .\)

What \(i s\) normal, when it comes to people's body temperatures? A random sample of 130 human body temperatures, provided by Allen Shoemaker \(^{11}\) in the Journal of Statistical Education, had a mean of \(98.25^{\circ}\) Fahrenheit and a standard deviation of \(0.73^{\circ}\) Fahrenheit. a. Construct a \(99 \%\) confidence interval for the average body temperature of healthy people. b. Does the confidence interval constructed in part a contain the value \(98.6^{\circ}\) Fahrenheit, the usual average temperature cited by physicians and others? If not, what conclusions can you draw?

Calculate the margin of error in estimating a binomial proportion \(p\) using samples of size \(n=100\) and the values of p given in Exercises \(15-19 .\) What value of p produces the largest margin of error? \(p=.5\)

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