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In a random sample of 200 people, 154 said that they watched educational television. Find the \(90 \%\) confidence interval of the true proportion of people who watched educational television. If the television company wanted to publicize the proportion of viewers, do you think it should use the \(90 \%\) confidence interval?

Short Answer

Expert verified
The 90% confidence interval is (0.7221, 0.8179). Yes, it should use it.

Step by step solution

01

Identify the Sample Proportion

Calculate the sample proportion, \( \hat{p} \), of people who watched educational television. The formula is \( \hat{p} = \frac{x}{n} \), where \( x \) is the number of successes (people who watched educational television) and \( n \) is the total sample size. Here, \( x = 154 \) and \( n = 200 \).
02

Calculate the Sample Proportion

Using the formula from Step 1, \( \hat{p} = \frac{154}{200} = 0.77 \). Thus, the sample proportion of people who watched educational television is 0.77.
03

Determine the Z-Score for 90% Confidence Level

For a 90% confidence interval, the Z-score is taken from the standard normal distribution table such that the central 90% of values lie within it. The Z-score corresponding to a 90% confidence level is approximately 1.645.
04

Compute the Standard Error

The standard error (SE) for the proportion is calculated using the formula: \( SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \). Substituting the known values, \( SE = \sqrt{\frac{0.77 \times (1 - 0.77)}{200}} \approx 0.0291 \).
05

Calculate the Margin of Error

The margin of error (ME) is given by multiplying the Z-score by the standard error: \( ME = Z \times SE \). Using the calculated values, \( ME = 1.645 \times 0.0291 \approx 0.0479 \).
06

Construct the Confidence Interval

The confidence interval (CI) is given by \( \hat{p} \pm ME \). So, the 90% confidence interval is \( 0.77 \pm 0.0479 \), which results in an interval of approximately (0.7221, 0.8179).
07

Interpret the Confidence Interval

This interval suggests that we are 90% confident that the true proportion of people who watch educational television lies between 72.21% and 81.79%.
08

Evaluate Publicity Decision

Using a 90% confidence interval is acceptable in this case, as it provides a reasonably precise estimate of the true proportion that balances accuracy with the width of the interval.

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Key Concepts

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

Sample Proportion
When dealing with proportions in statistics, the sample proportion is a foundation concept. It is essentially the fraction of the sample that exhibits a particular trait or outcome. In this exercise, we are interested in how many people out of a surveyed group watch educational television.

The formula to compute the sample proportion, denoted as \( \hat{p} \), is:
  • \( \hat{p} = \frac{x}{n} \)
Here, \( x \) represents the number of people who watch educational TV, and \( n \) stands for the total number of people in the sample. By applying this formula with our numbers, \( x = 154 \) and \( n = 200 \), we find that \( \hat{p} = 0.77 \).

This means that 77% of the sampled individuals are viewers of educational television, providing us with initial insight into viewer habits in this population.
Z-Score
The Z-score is a critical value derived from the standard normal distribution, and it enables us to estimate the confidence interval of our sample proportion.

For a confidence interval, the Z-score helps determine how "confident" you are that a true parameter lies within your calculated interval. In the context of a 90% confidence interval, the Z-score represents the point where central 90% of our data lies within. From statistical tables, the Z-score for 90% confidence is approximately 1.645.

This specific score allows us to balance precision with the range, ensuring that our estimate for the television viewership proportion is as accurate as possible without being overly narrow.
Margin of Error
Margin of Error (ME) represents the range around our sample proportion within which we expect the true population parameter to fall. It acts as a buffer for the interval.

The margin of error is calculated by multiplying the Z-score by the Standard Error (SE) of the proportion. The formula appears as follows:
  • \( ME = Z \times SE \)
For our calculations, the Margin of Error is found by multiplying the Z-score 1.645 by the SE, approximately 0.0291, resulting in a ME of roughly 0.0479.

This implies that our estimate of the sample proportion could deviate by about 4.79% above or below, explaining the possible variability in sampling.
Standard Error
Standard Error (SE) quantifies the variation of your sample proportion owing to sampling variability. It is crucial for constructing confidence intervals.

The standard error is calculated using this formula:
  • \( SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \)
This formula integrates our sample proportion's variance by multiplying \( \hat{p} \) and \( 1-\hat{p} \), then dividing by the sample size \( n \). For our scenario, substituting \( \hat{p} \) with 0.77 and \( n \) with 200 yields SE approximately 0.0291.

This small standard error signifies high precision in our sampling process, reinforcing the reliability of the population proportion estimation.

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Most popular questions from this chapter

A random sample of 50 four-year-olds attending day care centers provided a yearly tuition average of \(\$ 3987\) and the population standard deviation of \(\$ 630\). Find the \(90 \%\) confidence interval of the true mean. If a day care center were starting up and wanted to keep tuition low, what would be a reasonable amount to charge?

Parking Meter Revenue A one-sided confidence interval can be found for a mean by using $$ \mu>\bar{X}-t_{\alpha} \frac{s}{\sqrt{n}} \quad \text { or } \quad \mu<\bar{X}+t_{\alpha} \frac{s}{\sqrt{n}} $$ where \(t_{\alpha}\) is the value found under the row labeled One tail. Find two one-sided \(95 \%\) confidence intervals of the population mean for the data shown, and interpret the answers. The data represent the daily revenues in dollars from 20 parking meters in a small municipality. $$ \begin{array}{rrrr} 2.60 & 1.05 & 2.45 & 2.90 \\ 1.30 & 3.10 & 2.35 & 2.00 \\ 2.40 & 2.35 & 2.40 & 1.95 \\ 2.80 & 2.50 & 2.10 & 1.75 \\ 1.00 & 2.75 & 1.80 & 1.95 \end{array} $$

The numbers of faculty at 32 randomly selected state-controlled colleges and universities with enrollment under 12.000 students are shown below. Use these data to estimate the mean number of faculty at all state-controlled colleges and universities with enrollment under 12,000 with \(92 \%\) confidence. Assume \(\sigma=165.1\). \(\begin{array}{lllllllll}211 & 384 & 396 & 211 & 224 & 337 & 395 & 121 & 356 \\\ 621 & 367 & 408 & 515 & 280 & 289 & 180 & 431 & 176 \\ 318 & 836 & 203 & 374 & 224 & 121 & 412 & 134 & 539 \\ 471 & 638 & 425 & 159 & 324 & & & & \end{array}\)

It is believed that \(25 \%\) of U.S. homes have a direct satellite television receiver. How large a sample is necessary to estimate the true population of homes that do with \(95 \%\) confidence and within 3 percentage points? How large a sample is necessary if nothing is known about the proportion?

What assumption must be made when computing confidence intervals for variances and standard deviations?

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