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Chapter 22: Problem 26

Students are reluctant to report cheating by other students. A student project put this question to an SRS of 172 undergraduates at a large university: "You witness two students cheating on a quiz. Do you go to the professor?" Only 19 answered Yes. \({ }^{20}\) Give a \(95 \%\) confidence interval for the proportion of all undergraduates at this university who would report cheating.

Short Answer

Expert verified
The 95% confidence interval for the proportion is approximately 0.0644 to 0.1566.

Step by step solution

01

Identify Sample Proportion

The first step is to calculate the sample proportion (\(\hat{p}\)) of students who would report cheating. Given that 19 out of 172 students said "Yes," the sample proportion is \( \hat{p} = \frac{19}{172} \).
02

Calculate Sample Proportion Value

Now, perform the division: \( \hat{p} = \frac{19}{172} \approx 0.1105 \). This means that the sample proportion of students who would report cheating is approximately 0.1105.
03

Compute Standard Error

The standard error (SE) can be calculated using the formula: \( SE = \sqrt{\frac{\hat{p} (1-\hat{p})}{n}} \). Substituting the known values: \( SE = \sqrt{\frac{0.1105 \times 0.8895}{172}} \approx 0.0235 \).
04

Determine Z-Score

For a 95% confidence interval, the Z-score is typically 1.96. This comes from the standard normal distribution table which approximates the middle 95% of the data.
05

Calculate Margin of Error

The margin of error (ME) is given by \( ME = Z \times SE \). Substituting the known values: \( ME = 1.96 \times 0.0235 \approx 0.0461 \).
06

Construct Confidence Interval

Finally, the 95% confidence interval is constructed as \( \hat{p} \pm ME \). This becomes \( 0.1105 \pm 0.0461 \), which results in the interval approximately from 0.0644 to 0.1566.

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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 by \( \hat{p} \), represents the fraction or percentage of a sample with a particular trait of interest. In this scenario, we're determining the proportion of students who would report cheating. We use the number of students who answer "Yes," divided by the total sample size.

Here, 19 students reported they'd report cheating, from a total of 172 respondents. So, the sample proportion is \( \hat{p} = \frac{19}{172} \approx 0.1105 \).

This value means that, in our sample, roughly 11.05% of students would notify the professor if they saw cheating. Remember, this number solely represents the sample, not the entire population.
Standard Error
The standard error (SE) helps quantify the variability or standard deviation of a sample proportion. It estimates how much the sample proportion might differ from the true population proportion.

We calculate the standard error using the formula:
  • \( SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \)
In this example, \( \hat{p} = 0.1105 \) and \( n = 172 \), so plugging these values into the formula gives us:
\( SE = \sqrt{\frac{0.1105 \times 0.8895}{172}} \approx 0.0235 \).

The smaller the standard error, the more precise our sample estimate is. It tells us how much the sample proportion is expected to fluctuate from sample to sample.
Margin of Error
The margin of error (ME) describes the range within which our sample proportion is likely to fall in relation to the true population proportion. This measure reflects the extent of uncertainty associated with the sample estimate.

The calculation involves multiplying the standard error by the appropriate Z-score for the confidence level, defined as:
  • \( ME = Z \times SE \)
In this case, the Z-score for a 95% confidence level is 1.96. Therefore:
\( ME = 1.96 \times 0.0235 \approx 0.0461 \).

With this margin of error, we can be reasonably certain that the true proportion of students willing to report cheaters falls within this range of the sample proportion.
Z-score
The Z-score plays a crucial role in translating how standard deviations relate to a sample or population. It informs us about the probability of a value occurring within a standard normal distribution.

When constructing confidence intervals, the Z-score helps set the range dictated by our desired level of confidence. For a 95% confidence interval, the Z-score is typically 1.96, which means we expect the true parameter to lie within this range 95% of the time.

The Z-score selection is based on the idea of encompassing the central 95% of a normal distribution. This boundary ensures we capture the typical variations we believe exist in the true population proportion. It's crucial because it directly affects the margin of error and, consequently, the width of the confidence interval.

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Most popular questions from this chapter

A Gallup Poll in November 2012 found that \(54 \%\) of the people in the sample said they wanted to lose weight. The poll's margin of error for \(95 \%\) confidence was \(4 \%\). This means that (a) the poll used a method that gets an answer within \(4 \%\) of the truth about the population \(95 \%\) of the time. (b) we can be sure that the percent of all adults who want to lose weight is between \(50 \%\) and \(58 \%\). (c) if Gallup takes another poll using the same method, the results of the second poll will lie between \(50 \%\) and \(58 \%\).

Resistance training is a popular form of conditioning aimed at enhancing sports performance and is widely used among high school, college, and professional athletes, although its use for younger athletes is controversial. A random sample of 4111 patients aged 8-30 admitted to U.S. emergency rooms with the Consumer Product Safety Commission code "weightlifting" were obtained. These injuries were classified as "accidental" if caused by dropped weight or improper equipment use. Of the 4111 weightlifting injuries, 1552 were classified as accidental. \({ }^{10}\) Give a \(90 \%\) confidence interval for the proportion of weightlifting injuries in this age group that were accidental. Follow the four-step process as illustrated in Example 22.4.

Some shrubs have the useful ability to resprout from their roots after their tops are destroyed. Fire is a particular threat to shrubs in dry climates because it can injure the roots as well as destroy the aboveground material. One study of resprouting took place in a dry area of Mexico. \({ }^{32}\) The investigators clipped the tops of samples of several species of shrubs. In some cases, they also applied a propane torch to the stumps to simulate a fire. Of 12 specimens of the shrub Krameria cytisoides, five resprouted after fire. Estimate with \(90 \%\) confidence the proportion of all shrubs of this species that will resprout after fire.

A husband and wife, Stan and Lucretia, share a digital music player that has a feature that randomly selects which song to play. A total of 2643 songs have been loaded into the player, some by Stan and the rest by Lucretia. They are interested in determining whether they have loaded different proportions of songs into the player. Suppose that when the player was in the random- selection mode, 27 of the first 40 songs selected were songs loaded by Lucretia. Let \(p\) denote the proportion of songs that were loaded by Lucretia. (a) State the null and alternative hypotheses to be tested. How strong is the evidence that Stan and Lucretia have loaded different proportions of songs into the player? Make sure to check the conditions for the use of this test. (b) Are the conditions for the use of the large-sample confidence interval met? If so, estimate with \(95 \%\) confidence the proportion of songs that were loaded by Lucretia.

A high-school teacher in a lowincome urban school in Worcester, Massachusetts, used cash incentives to encourage student learning in his AP statistics class. \({ }^{28}\) In 2010,15 of the 61 students enrolled in his class scored a 5 on the AP statistics exam. Worldwide, the proportion of students who scored a 5 in 2010 was \(0.15\). Is this evidence that the proportion of students who would score a 5 on the AP statistics exam when taught by the teacher in Worcester using cash incentives is higher than the worldwide proportion of \(0.15\) ? (a) State hypotheses, find the \(P\)-value, and give your conclusions in the context of the problem. Do you have any reservations about using the \(z\) test for proportions for this data? (b) Does this study provide evidence that cash incentives cause an increase in the proportion of \(5 \mathrm{~s}\) on the AP statistics exam? Explain your answer.
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