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Chapter 7: Problem 31

According to The Washington Post, \(72 \%\) of high school seniors have a driver's license. Suppose we take a random sample of 100 high school seniors and find the proportion who have a driver's license. a. What value should we expect for our sample proportion? b. What is the standard error? c. Use your answers to parts a and \(\mathrm{b}\) to complete this sentence: We expect _____% to have their driver鈥檚 license, give or take _____%. d. Suppose we increased the sample size from 100 to 500 . What effect would this have on the standard error? Recalculate the standard error to see if your prediction was correct.

Short Answer

Expert verified
a) Expected sample proportion is \(72\%\). \n b) Standard error is \(4.47\%\). \n c) We expect \(72\%\) to have their driver鈥檚 license, give or take \(4.47\%\). \n d) The standard error would decrease with the increase in sample size. For a sample size of 500, the standard error is \(0.02\) or \(2\%\).

Step by step solution

01

Compute the Expected Proportion

The expected value for our sample proportion can be calculated directly from the given data. Since we know that \(72\%\) or \(0.72\) of high school seniors have a driver's license, we would expect the sample proportion to also be \(0.72\). Hence, the expected value is \(0.72\) or \(72\%\).
02

Calculate Standard Error

The standard error (SE) is used to measure the statistical accuracy of an estimate. In case of proportion it can be calculated by using the formula \(SE = \sqrt{{p(1-p)}/n}\), where \(p\) is the proportion (in our case that is \(0.72\)) and \(n\) is the sample size (which is 100 in our case). After plugging these values into the formula, we get \(SE = \sqrt{{0.72 * (1 - 0.72)}/100} = 0.0447\).
03

Complete the Sentence

With the above result, we can complete the sentence as follows: We expect \(72\%\) to have their driver's license, give or take \(4.47\%\).
04

Determine the Impact of Increasing the Sample Size

Increasing the sample size generally reduces the standard error, which in turn increases the precision of our sample estimate. Now, let's recalculate the standard error with \(n = 500\). Using the same formula as before, \(SE = \sqrt{{0.72 * (1 - 0.72)}/500} = 0.02\). As expected, the standard error decreased when we increased the sample size from 100 to 500.

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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 represents the percentage of a subset, in this case, high school seniors who possess a driver's license. When dealing with problems like these, the sample proportion serves as an estimate of the entire population's proportion. In our original exercise, it was given that 72% of high school seniors have a driver's license. If we randomly select 100 seniors, we anticipate finding approximately 72 who have a license, making our sample proportion 0.72 or 72%. This value is our best guess at the population proportion based on the sample.
Standard Error
The standard error (SE) is a measure used to understand the variability of a sample proportion as an estimate of a population proportion. It informs us about the accuracy of our estimate.
We calculate the SE using the formula: \( SE = \sqrt{\frac{p(1-p)}{n}} \)where:
  • \( p \) is the sample proportion, 0.72 in this case.
  • \( n \) is the sample size, 100 for our initial calculation.
After performing the calculation, for a sample size of 100, we find an SE of 0.0447. This means our estimate of 72% is accurate to within roughly 4.47 percentage points.
Sample Size
Sample size is the number of individuals observed in a survey or experiment. It plays a critical role in determining the precision of our sample estimate. Larger sample sizes generally produce more accurate estimates.
Increasing the sample size results in a lower standard error, which indicates a more precise estimate. For example, when the sample size increased from 100 to 500 in the exercise, the standard error decreased from 0.0447 to 0.02. This reduction illustrates heightened accuracy, suggesting our sample proportion is closer to the true population proportion.
Statistical Accuracy
Statistical accuracy refers to how close an estimate approximates the true population parameter. This concept is tightly linked with both the sample size and the standard error.
A lower standard error implies higher statistical accuracy because it shows less variability in repeated samples. This enhances our confidence in the sample proportion as a valid representation of the true population proportion.
Thus, by increasing the sample size from 100 to 500, the standard error decreased, leading to improved statistical accuracy. Such a change ensures that our 72% estimate is a much tighter and reliable approximation of the population proportion.

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Most popular questions from this chapter

Bob Ross hosted a weekly television show, The Joy of Painting, on PBS in which he taught viewers how to paint. During each episode, he produced a complete painting while teaching viewers how they could produce a similar painting. Ross completed 30,000 paintings in his lifetime. Although it was an art instruction show, PBS estimated that only \(10 \%\) of viewers painted along with Ross during his show based on surveys of viewers. For each of the following, also identify the population and explain your choice. a. Is the number 30,000 a parameter or a statistic? b. Is the number \(10 \%\) a parameter or a statistic?

You need to select a simple random sample of four from eight friends who will participate in a survey. Assume the friends are numbered \(1,2,3,4,5\), 6,7, and \(8 .\) Select four friends, using the two lines of numbers in the next column from a random number table. Read off each digit, skipping any digit not assigned to one of the friends. The sampling is without replacement, meaning that you cannot select the same person twice. Write down the numbers chosen. The first person is number 7 . $$ \begin{array}{lll} 07033 & 75250 & 34546 \\ \hline 75298 & 33893 & 64487 \end{array} $$ Which four friends are chosen?

A recent Monmouth University poll found that 675 out of 1008 randomly selected people in the United States felt that college and universities with big sports programs placed too much emphasis on athletics over academics. Assuming the conditions for using the CLT were met, use the Minitab output provided to answer these questions. $$ \begin{aligned} &\text { Descriptive Statistics }\\\ &\begin{array}{rrrr} \mathrm{N} & \text { Event } & \text { Sample } \mathrm{p} & 95 \% \mathrm{Cl} \text { for } \mathrm{p} \\ \hline 1008 & 675 & 0.669643 & (0.639648,0.698643) \end{array} \end{aligned} $$ a. Complete this sentence: I am \(95 \%\) confident that the population proportion believing that colleges and universities with big sports programs place too much emphasis on athletics over academics is between _____ and _____. Report each number as a percentage rounded to one decimal place. b. Suppose a sports blogger wrote an article claiming that the majority of Americans believe that colleges and university with big sports programs place too much emphasis on athletics over academics. Does this confidence interval support the blogger's claim? Explain your reasoning.

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.

A Harris poll asked Americans in 2016 and 2017 if they were happy. In \(2016,31 \%\) reported being happy and in 2017 , \(33 \%\) reported being happy. Assume the sample size for each poll was 1000 . A \(95 \%\) confidence interval for the difference in these proportions \(p_{1}-p_{2}\) (where proportion 1 is proportion happy in 2016 and proportion 2 is the proportion happy in 2017) is \((-0.06,0.02)\). Interpret this confidence interval. Does the interval contain \(0 ?\) What does this tell us about happiness among American in 2016 and \(2017 ?\)
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