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In order to estimate the proportion of entering students who graduate within six years, the administration at a state university examined the records of 600 randomly selected students who entered the university six years ago, and found that 312 had graduated. a. Give a point estimate of the six-year graduation rate, the proportion of entering students who graduate within six years. b. Assuming that the sample is sufficiently large, construct a \(98 \%\) confidence interval for the six-year graduation rate.

Short Answer

Expert verified
Point estimate is 0.52, and the 98% confidence interval is [0.4727, 0.5673].

Step by step solution

01

Determine the Point Estimate

A point estimate of the proportion of students who graduated within six years is calculated by dividing the number of students who graduated by the total number of students examined. Here, 312 students graduated out of 600, so the point estimate \( \hat{p} = \frac{312}{600} \). Calculate this to find \( \hat{p} = 0.52 \).
02

Calculate the Standard Error (SE)

The standard error of the proportion is given by the formula \( SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \), where \( n \) is the sample size. Therefore, \( SE = \sqrt{\frac{0.52(1-0.52)}{600}} \). Calculate this to find \( SE \approx 0.0203 \).
03

Find the Z-Score for a 98% Confidence Interval

For a 98% confidence interval, we use the Z-score corresponding to 98%. The Z-score, often found from a Z-table or standard normal distribution table, is approximately 2.33 for a 98% confidence interval.
04

Calculate the Margin of Error (ME)

The margin of error is determined by multiplying the Z-score by the standard error: \( ME = Z \times SE \). Therefore, \( ME = 2.33 \times 0.0203 \approx 0.0473 \).
05

Construct the Confidence Interval

The confidence interval is \( \hat{p} \pm ME \). Substituting the values: \( 0.52 \pm 0.0473 \). Therefore, the 98% confidence interval is approximately \([0.4727, 0.5673]\).

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Key Concepts

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

Point Estimate
To estimate a population parameter, we often use a point estimate. A point estimate is a single value that serves as a "best guess" for a parameter of interest. In the context of proportions, it is like drawing a direct line from the sample to the population.

The given problem is about finding the graduation rate based on a sample of students. Here, the proportion of students who graduate within six years is represented as the point estimate \( \hat{p} \). This is calculated as follows:
  • Divide the number of successful cases, or graduates in this instance (312 students), by the total number of cases (600 students).
The point estimate is computed as: \[ \hat{p} = \frac{312}{600} = 0.52 \] This tells us that 52% of the students in our sample graduated within six years. This proportion serves as a straightforward estimate of the actual graduation rate in the whole population of students entering the university.
Standard Error
The standard error is another significant concept when discussing statistics. It measures the variability or dispersion of a sample statistic and helps us understand how much the point estimate might differ from the true population parameter.

To find the standard error of the proportion, we use the formula:
  • \( SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \)
  • Where \( \hat{p} \) is the point estimate, and \( n \) is the sample size.
For our example, the standard error is calculated as: \[ SE = \sqrt{\frac{0.52 \times (1 - 0.52)}{600}} = 0.0203 \] This value of approximately 0.0203 indicates the standard deviation of the sample proportion estimate. It shows how much the sample proportion (point estimate) is expected to vary from one sample to another if the experiment is repeated numerous times.
Z-Score
The Z-score is a statistical measurement that describes a value's relation to the mean of a group of values, measured in terms of standard deviations from the mean. It is a critical component when constructing confidence intervals, especially for large samples.

For confidence intervals, the Z-score corresponds to the level of confidence we aim to achieve. The Z-score assures us how many standard deviations away a certain data point is from the mean:
  • A Z-score for a 98% confidence level, for instance, is about 2.33.
This value implies that there is a 98% probability that the point estimate lies within 2.33 standard deviations of the true parameter value.
Margin of Error
The margin of error helps to create a range around the point estimate, showing how much we can expect the true parameter to differ from our estimate. It is especially useful in survey results to quantify the uncertainty.

To calculate the margin of error, we multiply the Z-score by the standard error:
  • \( ME = Z \times SE \)
In the scenario of the graduation rate, this becomes: \[ ME = 2.33 \times 0.0203 \approx 0.0473 \] This margin of error tells us the range around our point estimate (52%) within which we can be 98% confident that the true proportion lies.
Finally, the confidence interval can be expressed as the point estimate plus and minus the margin of error. For instance, a 98% confidence interval for the graduation rate is roughly \([0.4727, 0.5673]\), indicating a high level of certainty that the real graduation rate falls within this range.

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Most popular questions from this chapter

Large Data Sets 4 and 4 Alist the results of 500 tosses of a die. Let \(p\) denote the proportion of all tosses of this die that would result in a four. Use the sample data to construct a \(90 \%\) confidence interval for \(p\).

A manufacturer of chokes for shotguns tests a choke by shooting 15 patterns at targets 40 yards away with a specified load of shot. The mean number of shot in a 30 -inch circle is 53.5 with standard deviation 1.6. Construct an \(80 \%\) confidence interval for the mean number of shot in a 30 -inch circle at 40 yards for this choke with the specified load. Assume a normal distribution of the number of shot in a 30 - inch circle at 40 yards for this choke.

In a random sample of 250 employed people, 61 said that they bring work home with them at least occasionally. a. Give a point estimate of the proportion of all employed people who bring work home with them at least occasionally. b. Construct a \(99 \%\) confidence interval for that proportion.

Confidence intervals constructed using the formula in this section often do not do as well as expected unless \(n\) is quite large, especially when the true population proportion is close to either 0 or \(1 .\) In such cases a better result is obtained by adding two successes and two failures to the actual data and then computing the confidence interval. This is the same as using the formula $$ \begin{aligned} &\tilde{p} \pm z_{\alpha} / 2 \sqrt{\frac{\hat{p}(1-\dot{p})}{\tilde{n}}}\\\ &\text { where }\\\ &\tilde{p}-\frac{x+2}{n+4} \operatorname{and} n-n+4 \end{aligned} $$ Suppose that in a random sample of 600 households, 12 had no telephone service of any kind. Use the adjusted confidence interval procedure just described to form a \(99.9 \%\) confidence interval for the proportion of all households that have no telephone service of any kind.

In order to estimate the speaking vocabulary of three-year-old children in a particular socioeconomic class, a sociologist studies the speech of four children. The mean and standard deviation of the sample are \(\bar{x}-1120\) and \(s=215\) words. Assuming that speaking vocabularies are normally distributed, construct an \(80 \%\) confidence interval for the mean speaking vocabulary of all three-year-old children in this socioeconomic group.

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