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In 2018 it was estimated that approximately \(45 \%\) of the American population watches the Super Bowl yearly. Suppose a sample of 120 Americans is randomly selected. After verifying the conditions for the Central Limit Theorem are met, find the probability that at the majority (more than \(50 \%\) ) watched the Super Bowl. (Source: vox.com).

Short Answer

Expert verified
Therefore, the probability that the majority (more than \(50\%\)) of the sampled Americans watched the Super Bowl is approximately \(0.1635\) or \(16.35\%.\)

Step by step solution

01

Define Population Proportion and Sample Size

From the problem, the population proportion, \(p\) is \(0.45\) and the sample size, \(n\) is \(120\). We are asked to find the probability that the sample proportion, \(\hat{p}\), is greater than \(0.50\).
02

Calculate the Mean and Standard Deviation of the Sample Proportion

The mean of the sample proportions, \(\mu_{\hat{p}}\) is equal to the population proportion, \(p\), and the standard deviation of the sample proportions, \(\sigma_{\hat{p}}\) can be calculated using the formula \(\sqrt{ \frac{p(1-p)}{n} }\). This gives us \(\mu_{\hat{p}} = 0.45\) and \(\sigma_{\hat{p}} = \sqrt{ \frac{0.45 \times 0.55}{120}} = 0.051\).
03

Calculate the z-score

The z-score is calculated with the formula \(z = \frac{\hat{p} - \mu_{\hat{p}}}{\sigma_{\hat{p}}}\). We wish to find the probability that \(\hat{p} > 50\%\). Thus, we will calculate the z-score at 50% (or 0.50) which gives us \(z = \frac{0.50 - 0.45}{0.051} = 0.98\).
04

Find the Probability

We need to find \(P(\hat{p} > 0.50)\), which can also be written as \(P(Z > 0.98)\). From standard normal tables, \(P(Z > 0.98)\) is approximately \(0.1635\). Alternately, 1 - \(P(Z < 0.98)\ = 1 - 0.8365 = 0.16350\).

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Key Concepts

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

Population Proportion
Understanding the concept of population proportion is essential when conducting surveys or experiments involving a large group. In statistics, the population proportion, denoted as p, represents the fraction of the population that has a certain characteristic. For instance, in the provided exercise, the population proportion is the estimated percentage of Americans who watch the Super Bowl, which is 45% or \(p = 0.45\).

When analyzing random samples from this population, such as the selected group of 120 Americans, we assume each individual independently has the same probability (\(p\)) of watching the Super Bowl. This is the basis for predicting the outcomes within the sample and determining the probability of a certain number of people who watched the event.
Sample Size
The sample size, denoted as n, is a pivotal factor in statistical analysis. It represents the number of individual observations used to estimate a characteristic of a population. In our exercise, the sample size is 120, which is sufficiently large for applying the Central Limit Theorem (CLT).

The size of the sample dictates the accuracy of our estimates; larger samples tend to produce more reliable and precise results. However, it's crucial to ensure that the samples are randomly selected to represent the population effectively. The chosen sample size impacts the calculation of the standard deviation of the sampling distribution, which in turn affects the z-score and the resulting probabilities.
Z-Score
The z-score is a statistical measure that describes a value's relationship to the mean of a group of values. It is measured in terms of standard deviations from the mean. In the context of the Central Limit Theorem, a z-score is used to find the probability that the sample proportion falls within a certain range.

For a given sample proportion, the z-score is calculated by the formula \(z = \frac{{\hat{p} - \mu_{\hat{p}}}}{{\sigma_{\hat{p}}}}\), where \(\hat{p}\) is the sample proportion, \(\mu_{\hat{p}}\) is the mean of the sample proportions, and \(\sigma_{\hat{p}}\) is the standard deviation. In simpler terms, it tells us how many standard deviations a data point (or a proportion in our case) is from the mean. This standardized score is crucial in determining the probability associated with the observed outcome.
Probability
The term probability in statistics refers to the measure of the likelihood that an event will occur. It is quantified as a number between 0 and 1, where 0 indicates impossibility, and 1 indicates certainty. The exercise requires us to find the probability that more than 50% of the sample size watches the Super Bowl, symbolically represented as \(P(\hat{p} > 0.50)\).

Using the Central Limit Theorem, we can assume that the sample proportions follow a normal distribution, which allows us to use the z-score to find the corresponding probability from the standard normal distribution. This becomes a vital step in hypothesis testing and making inferences about the overall population based on the sample data.
Standard Deviation
The concept of standard deviation plays a significant role in statistics, particularly when applying the Central Limit Theorem. It measures the spread or dispersion of a set of values. In our context, the standard deviation of the sample proportion (\(\sigma_{\hat{p}}\)) helps us understand how much variation there is from the average (mean) proportion of the population. A lower standard deviation indicates that the values tend to be closer to the mean, while a larger standard deviation indicates that the values are spread out over a wider range.

In statistical formulas, standard deviation is often denoted as \(\sigma\), with the standard deviation of the sample proportion found using the formula \(\sqrt{\frac{{p(1-p)}}{n}}\). This calculated value is crucial for determining the z-score, which is then used to find probabilities for specific sample observations.

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Most popular questions from this chapter

In 2016 and 2017 Gallup asked American adults about their amount of trust they had in the judicial branch of government. In \(2016,61 \%\) expressed a fair amount or great deal of trust in the judiciary. In \(2017,68 \%\) of Americans felt this way. These percentages are based on samples of 1022 American adults. a. Explain why it would be inappropriate to conclude, based on these percentages alone, that the percentage of Americans who had a fair amount or great deal of trust in the judicial branch of government increased from 2016 to 2017 . b. Check that the conditions for using a two-proportion confidence interval hold. You can assume that the sample is a random sample. c. Construct a \(95 \%\) confidence interval for the difference in the proportions of Americans who expressed this level of trust in the judiciary, \(p_{1}-p_{2}\), where \(p_{1}\) is the 2016 population proportion and \(p_{2}\) is the 2017 population proportion. d. Based on the confidence interval constructed in part c, can we conclude that proportion of Americans with this level of trust in the judiciary increased from 2016 to \(2017 ?\) Explain.

The website scholarshipstats.com collected data on all 5341 NCAA basketball players for the 2017 season and found a mean height of 77 inches. Is the number 77 a parameter or a statistic? Also identify the population and explain your choice.

Suppose that, when taking a random sample of three students' GPAs, you get a sample mean of \(3.90 .\) This sample mean is far higher than the collegewide (population) mean. Does that prove that your sample is biased? Explain. What else could have caused this high mean?

In carrying out a study of views on capital punishment, a student asked a question two ways: 1\. With persuasion: "My brother has been accused of murder and he is innocent. If he is found guilty, he might suffer capital punishment. Now do you support or oppose capital punishment?" 2\. Without persuasion: "Do you support or oppose capital punishment?" Here is a breakdown of her actual data. $$ \begin{aligned} &\text { Men }\\\ &\begin{array}{lcc} & \begin{array}{c} \text { With } \\ \text { persuasion } \end{array} & \begin{array}{c} \text { No } \\ \text { persuasion } \end{array} \\ \hline \text { For capital punishment } & 6 & 13 \\ \hline \text { Against capital punishment } & 9 & 2 \\ \text { Women } \end{array}\\\ &\begin{array}{lcc} & \begin{array}{c} \text { With } \\ \text { persuasion } \end{array} & \begin{array}{c} \text { No } \\ \text { persuasion } \end{array} \\ \hline \text { For capital punishment } & 2 & 5 \\ \hline \text { Against capital punishment } & 8 & 5 \end{array} \end{aligned} $$ a. What percentage of those persuaded against it support capital punishment? b. What percentage of those not persuaded against it support capital punishment? c. Compare the percentages in parts a and b. Is this what you expected? Explain.

From Formula 7.2, an estimate for margin of error for a \(95 \%\) confidence interval is \(m=2 \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\) where \(\mathrm{n}\) is the required sample size and \(\hat{p}\) is the sample proportion. Since we do not know a value for \(\hat{p}\), we use a conservative estimate of \(0.50\) for \(\hat{p}\). Replace \(\hat{p}\) with \(0.50\) in the formula and simplify.

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