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Chapter 8: Problem 57

Movie recommendation In a quick poll at the exit of a movie theater, 8 out of 12 randomly polled viewers said they would recommend the movie to their friends. a. Construct an appropriate \(95 \%\) confidence interval for the population proportion. b. Is it plausible that only half of all the viewers will be willing to recommend the film to their friends? Explain.

Short Answer

Expert verified
a. (0.400, 0.934) b. Yes, 0.5 is within the interval.

Step by step solution

01

Identify the sample proportion

From the scenario, 8 out of 12 viewers recommend the movie. Therefore, the sample proportion \( \hat{p} \) can be calculated as \( \hat{p} = \frac{8}{12} = \frac{2}{3} \approx 0.667 \).
02

Identify the critical value for 95% confidence interval

For a 95% confidence level, the critical value (z-score) is approximately 1.96 when using a standard normal distribution. This value is common for two-tailed tests.
03

Calculate the standard error

The standard error (SE) for the sample proportion can be calculated using the formula: \( SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \), where \( n = 12 \). So, \( SE = \sqrt{\frac{0.667 \times 0.333}{12}} \approx 0.136 \).
04

Compute the confidence interval

The formula for the confidence interval is \( \hat{p} \pm z \times SE \). Thus, the interval is \( 0.667 \pm 1.96 \times 0.136 \approx 0.667 \pm 0.267 \). This gives a confidence interval of approximately (0.400, 0.934).
05

Evaluate if half of the viewers recommending the movie is plausible

We need to check if a proportion of 0.5 (or 50%) is within our calculated confidence interval (0.400, 0.934). Since 0.5 is within this range, it is plausible that 50% of viewers would recommend the movie.

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Key Concepts

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

Sample Proportion
The sample proportion, often denoted as \( \hat{p} \), is a vital statistic that helps us understand the behavior of a specific characteristic within a subset of the population. In this example, the sample proportion is derived from 8 out of 12 viewers who recommended the movie, calculated as \( \hat{p} = \frac{8}{12} = \frac{2}{3} \approx 0.667 \). This fraction represents the likelihood of a single randomly chosen individual from the sample endorsing the film.

Calculating the sample proportion is straightforward: simply divide the number of favorable outcomes by the total number of observations. It's crucial because it forms the basis of statistical analysis, allowing us to infer the larger population's behavior regarding this trait.
Critical Value
The critical value in statistical procedures, especially in the context of confidence intervals, refers to the point on the standard normal distribution curve that corresponds to the specified level of confidence. Normally, for a 95% confidence interval, the critical value (z-score) is 1.96. This value plays a key role in determining the margin of error.

Critical values are essential in hypothesis testing as well, acting as a threshold to determine whether or not to reject a hypothesis. Selecting the correct critical value ensures that our interval or hypothesis retains the desired confidence level. It provides a boundary for how much random sampling error can impact the results.
Standard Error
The standard error (SE) measures the statistical accuracy of a sample proportion by indicating how much this proportion is expected to vary from the true population proportion if we repeatedly sampled the population. It is calculated using the formula: \[ SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \]where \( \hat{p} \) is the sample proportion, and \( n \) is the sample size. For our example, the standard error is about 0.136.

A smaller standard error indicates that the sample proportion is a more accurate estimate of the true population proportion. Therefore, increasing the sample size or obtaining a sample proportion closer to 0.5 can reduce the standard error, making estimates more reliable. This metric is crucial for constructing confidence intervals, directly influencing their width.
Population Proportion
The population proportion represents the fraction of the entire population that shares a specific characteristic, in this case, those who would recommend the film. It is denoted as \( p \), which we estimate through a sample proportion \( \hat{p} \).

We never know the exact value of the population proportion, but we can estimate it using the confidence interval derived from the sample proportion. For instance, the computed confidence interval of approximately (0.400, 0.934) suggests that the true population proportion falls somewhere within this range with 95% confidence.

Understanding the concept of population proportion is beneficial as it allows statisticians to make wider inferences about the population based on limited data. It is essentially the backbone of inference statistics, helping translate sample findings into broader population insights.

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Most popular questions from this chapter

In a clinical study (the same as mentioned in Example 4 ), 3900 subjects were vaccinated with a vaccine manufactured by growing cells in fertilized chicken eggs. Over a period of roughly 28 weeks, 24 of these subjects developed the flu. a. Find the point estimate of the population proportion that were vaccinated with the vaccine but still developed the flu. b. Find the standard error of this estimate. c. Find the margin of error for a \(95 \%\) confidence interval. d. Construct the \(95 \%\) confidence interval for the population proportion. Interpret the interval. e. Can you conclude that fewer than \(1 \%\) of all people vaccinated with the vaccine will develop the flu? Explain by using results from part d.

Web survey to get large \(n\) A newspaper wants to gauge public opinion about legalization of marijuana. The sample size formula indicates that it need a random sample of 875 people to get the desired margin of error. But surveys cost money, and it can only afford to randomly sample 100 people. Here's a tempting alternative: If it places a question about that issue on its website, it will get more than 1000 responses within a day at little cost. Is it better off with the random sample of 100 responses or the website volunteer sample of more than 1000 responses? (Hint: Think about the issues discussed in Section 4.2 about proper sampling of populations.)

Wage discrimination? According to a union agreement, the mean income for all senior-level assembly-line workers in a large company equals \(\$ 500\) per week. A representative of a women's group decides to analyze whether the mean income for female employees matches this norm. For a random sample of nine female employees, using software, she obtains a \(95 \%\) confidence interval of ( 371 , 509 ). Explain what is wrong with each of the following interpretations of this interval. a. We infer that \(95 \%\) of the women in the population have income between \(\$ 371\) and \(\$ 509\) per week. b. If random samples of nine women were repeatedly selected, then \(95 \%\) of the time the sample mean income would be between \(\$ 371\) and \(\$ 509\). c. We can be \(95 \%\) confident that \(\bar{x}\) is between \(\$ 371\) and \(\$ 509 .\) d. If we repeatedly sampled the entire population, then \(95 \%\) of the time the population mean would be between \(\$ 371\) and \(\$ 509 .\)

Google Glass Google started selling Google Glass (a type of wearable technology that projects information onto the eye) in the United States in spring 2014 for \(\$ 1500\). A national poll reveals that out of 500 people sampled, nobody owned Google Glass. a. Find the sample proportion who don't own Google Glass and its standard error. b. Find a \(95 \%\) confidence interval, using the large-sample formula. Is it sensible to conclude that no one in the United States owns Google Glass?] c. Why is it not appropriate to use the ordinary large-sample confidence interval in part b? Use a more appropriate approach and interpret the result. d. Is it plausible to say that fewer than \(1 \%\) of the population own Google Glass? Explain.

In a 2014 Consumer Reports article titled, "The High Cost of Cheap Chicken," the magazine reported that out of 316 chicken breasts bought in retail stores throughout the United States, 207 contained E. coli bacteria. a. Find and interpret a \(99 \%\) confidence interval for the population proportion of chicken breasts that contain E. coli. Can you conclude that the proportion of chicken containing E. coli exceeds \(50 \%\) ? Why? b. Without doing any calculation, explain whether the interval in part a would be wider or narrower than a \(95 \%\) confidence interval for the population proportion.
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