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Chapter 9: Problem 18

Consider taking a random sample from a population with \(p=0.40\). a. What is the standard error of \(\hat{p}\) for random samples of size \(100 ?\) b. Would the standard error of \(\hat{p}\) be greater for samples of size 100 or samples of size \(200 ?\) c. If the sample size were doubled from 100 to 200 , by what factor would the standard error of \(\hat{p}\) decrease?

Short Answer

Expert verified
a. The standard error of \(\hat{p}\) for random samples of size 100 is 0.049. b. The standard error of \(\hat{p}\) is greater for samples of size 100 than samples of size 200. c. The standard error of \(\hat{p}\) decreases by a factor of approximately 1.416 when the sample size is doubled from 100 to 200.

Step by step solution

01

Identify the given values

In this exercise, we are given the population proportion \(p = 0.40\) and the sample size \(n=100\).
02

Calculate the standard error of \(\hat{p}\) for the sample size of 100

Using the standard error formula and the given values, we can calculate the standard error of \(\hat{p}\) for a sample size of 100: \[SE(\hat{p}) = \sqrt{\frac{0.40(1-0.40)}{100}} = \sqrt{\frac{0.40(0.60)}{100}} = \sqrt{\frac{0.24}{100}} = 0.049\] The standard error of \(\hat{p}\) for random samples of size \(100\) is \(0.049\). #b. Comparing standard errors for sample sizes 100 and 200#
03

Calculate the standard error of \(\hat{p}\) for a sample size of 200

Using the standard error formula and the given population proportion, we can calculate the standard error of \(\hat{p}\) for a sample size of 200: \[SE(\hat{p}) = \sqrt{\frac{0.40(1-0.40)}{200}} = \sqrt{\frac{0.40(0.60)}{200}} = \sqrt{\frac{0.24}{200}} = 0.0346\] The standard error of \(\hat{p}\) for random samples of size \(200\) is \(0.0346\).
04

Compare the standard errors for sample sizes 100 and 200

Comparing the standard errors for sample sizes 100 and 200, we observe that: \(SE(\hat{p})_{n=100} = 0.049\) and \(SE(\hat{p})_{n=200} = 0.0346\). Since \(0.0346 < 0.049\), the standard error of \(\hat{p}\) is greater for samples of size \(100\) than samples of size \(200\). #c. Finding the factor by which the standard error decreases when doubling the sample size#
05

Calculate the ratio of the standard errors for sample sizes 100 and 200

To find the factor by which the standard error decreases when doubling the sample size, we calculate the ratio of the standard errors for sample sizes 100 and 200: \[ \text{Factor} = \frac{SE(\hat{p})_{n=100}}{SE(\hat{p})_{n=200}} = \frac{0.049}{0.0346} = 1.416\] The factor by which the standard error of \(\hat{p}\) decreases when the sample size is doubled from 100 to 200 is approximately \(1.416\).

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Key Concepts

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

Population Proportion
Understanding the population proportion is crucial when working with statistics, especially in the context of sampling. The population proportion, denoted by \(p\), reflects the percentage of a particular outcome within an entire population. For example, if we want to know the proportion of left-handed students in a school, and 40 out of 100 students are left-handed, the population proportion \(p\) would be 0.40 or 40%.

When conducting a study, it's often impractical or impossible to examine the entire population, so researchers take a sample and use the sample proportion, denoted by \(\hat{p}\), to estimate \(p\). It's important to note the difference: \(p\) represents the true proportion in the entire population, while \(\hat{p}\) represents the proportion in a sample, which serves as an estimate of the true proportion. Despite being an estimate, \(\hat{p}\) becomes more reflective of \(p\) as the sample size increases, under the law of large numbers.
Sample Size
Sample size, symbolized as \(n\), refers to the number of observations or individuals selected from the population to be included in a sample. It plays a significant role in the reliability of statistical results; the larger the sample size, the more confident researchers can be that their sample proportion \(\hat{p}\) accurately represents the population proportion \(p\).

A small sample might be skewed by outliers or not representative of the population, leading to potential bias. A larger sample helps to mitigate this by capturing a broader range of data, which is particularly important when aiming for high precision in estimates.

In the original exercise, we compared sample sizes of 100 and 200 to observe how they affect the standard error of \(\hat{p}\). A larger sample size resulted in a smaller standard error, signifying that the estimate from the larger sample is likely to be closer to the true population proportion.
Standard Error Calculation
The standard error (SE) measures the variability of a statistic鈥攕uch as the sample proportion \(\hat{p}\)鈥攆rom sample to sample. It gives us an idea of how precise our estimate of the population proportion \(p\) is. The formula for the standard error of the sample proportion is:

\[ SE(\hat{p}) = \sqrt{\frac{p(1-p)}{n}} \]
where \(p\) is the population proportion and \(n\) is the sample size. Coming back to the initial example with \(p=0.40\) and \(n=100\), we used this formula to calculate the standard error, which turned out to be 0.049.

Through the calculation, we see that the standard error is inversely related to the square root of the sample size; as the sample size increases, the standard error decreases. Doubling the sample size from 100 to 200 decreased the standard error by a factor of roughly 1.416. This principle forms a key part of determining how large a sample we need to have a certain level of confidence in our estimates.

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Most popular questions from this chapter

Appropriate use of the interval $$ \hat{p} \pm(z \text { critical value }) \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} $$ requires a large sample. For each of the following combinations of \(n\) and \(\hat{p}\), indicate whether the sample size is large enough for this interval to be appropriate. a. \(n=100\) and \(\hat{p}=0.70\) b. \(n=40\) and \(\hat{p}=0.25\) c. \(n=60\) and \(\hat{p}=0.25\) d. \(n=80\) and \(\hat{p}=0.10\)

In a survey of 800 college students in the United States, 576 indicated that they believe that a student or faculty member on campus who uses language considered racist, sexist, homophobic, or offensive should be subject to disciplinary action ("Listening to Dissenting Views Part of Civil Debate," USA TODAY, November 17,2015 ). Assuming that the sample is representative of college students in the United States, construct and interpret a \(95 \%\) confidence interval for the proportion of college students who have this belief.

The report "The 2016 Consumer Financial Literacy Survey" (The National Foundation for Credit Counseling, www.nfcc.org, retrieved October 28,2016 ) summarized data from a representative sample of 1668 adult Americans. Based on data from this sample, it was reported that over half of U.S. adults would give themselves a grade of \(\mathrm{A}\) or \(\mathrm{B}\) on their knowledge of personal finance. This statement was based on observing that 934 people in the sample would have given themselves a grade of \(\mathrm{A}\) or \(\mathrm{B}\). a. Construct and interpret a \(95 \%\) confidence interval for the proportion of all adult Americans who would give themselves a grade of \(\mathrm{A}\) or \(\mathrm{B}\) on their financial knowledge of personal finance. b. Is the confidence interval from Part (a) consistent with the statement that a majority of adult Americans would give themselves a grade of \(\mathrm{A}\) or \(\mathrm{B}\) ? Explain why or why not.

In spite of the potential safety hazards, some people would like to have an Internet connection in their car. A preliminary survey of adult Americans has estimated the proportion of adult Americans who would like Internet access in their car to be somewhere around 0.30 (USA TODAY, May 1 , 2009). Use the given preliminary estimate to determine the sample size required to estimate this proportion with a margin of error of 0.02

Suppose that a city planning commission wants to know the proportion of city residents who support installing streetlights in the downtown area. Two different people independently selected random samples of city residents and used their sample data to construct the following confidence intervals for the population proportion: Interval 1:(0.28,0.34) Interval 2:(0.31,0.33) (Hint: Consider the formula for the confidence interval given on page 444.) a. Explain how it is possible that the two confidence intervals are not centered in the same place. b. Which of the two intervals conveys more precise information about the value of the population proportion? c. If both confidence intervals have a \(95 \%\) confidence level, which confidence interval was based on the smaller sample size? How can you tell? d. If both confidence intervals were based on the same sample size, which interval has the higher confidence level? How can you tell?
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