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Chapter 7: Problem 40

Suppose it is known that \(60 \%\) of employees at a company use a Flexible Spending Account (FSA) benefit. a. If a random sample of 200 employees is selected, do we expect that exactly \(60 \%\) of the sample uses an FSA? Why or why not? b. Find the standard error for samples of size 200 drawn from this population. What adjustments could be made to the sampling method to produce a sample proportion that is more precise?

Short Answer

Expert verified
a) No, we don't expect that exactly \(60\%\) of the sample will use a FSA. This is because in random sampling, the sample proportion can vary due to sampling variability. b) The standard error for this sample is calculated by the formula \(\sqrt{(p*(1-p)/n)}\), resulting in a specific value. To improve sample proportion precision, we could increase the sample size, ensure random selection of the sample and ensure that the sample is representative of the overall population.

Step by step solution

01

Understand Proportions

Start by understanding the concept of proportions. A population proportion is the fraction or percentage of a group possessing a certain characteristic (in this case, \(60\%\) of the employees using a FSA). Sample proportion, on the other hand, refers to the fraction or percentage of our sample that possesses that characteristic. When we select a sample randomly, we expect its proportion to be close to the population proportion, but not precisely the same. This is due to the variability inherent in random sampling.
02

Calculate the Standard Error

Next, calculate the standard error, which quantifies the expected variability in sample proportions. The formula for standard error for proportions is \(\sqrt{(p*(1-p)/n)}\), where \(p\) is the population proportion and \(n\) is the sample size. For our problem, \(p = 0.60\) and \(n = 200\). Thus, the standard error \(= \sqrt{(0.60*(1-0.60)/200)}\).
03

Improve the Precision of Sample Proportion

The precision of a sample proportion can be improved by increasing the sample size. Larger samples tend to produce statistics that are closer to the value of the population parameter, because they are less influenced by variability due to random chance. Sampling methodology can also be improved by ensuring that samples are selected randomly and are representative of the overall population.

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Most popular questions from this chapter

A 2017 Gallup poll reported that 658 out of 1028 U.S. adults believe that marijuana should be legalized. When Gallup first polled U.S. adults about this subject in 1969 , only \(12 \%\) supported legalization. Assume the conditions for using the CLT are met. a. Find and interpret a \(99 \%\) confidence interval for the proportion of U.S. adults in 2017 that believe marijuana should be legalized. b. Find and interpret a \(95 \%\) confidence interval for this population parameter. c. Find the margin of error for each of the confidence intervals found in parts a and b. d. Without computing it, how would the margin of error of a \(90 \%\) confidence interval compare with the margin of error for the \(95 \%\) and \(99 \%\) intervals? Construct the \(90 \%\) confidence interval to see if your prediction was correct.

A school district conducts a survey to determine whether voters favor passing a bond to fund school renovation projects. All registered voters are called. Of those called, \(15 \%\) answer the survey call. Of those who respond, \(62 \%\) say they favor passing the bond. Give a reason why the school district should be cautious about predicting that the bond will pass.

According to a 2017 Pew Research report, \(40 \%\) of millennials have a BA degree. Suppose we take a random sample of 500 millennials and find the proportion who have a BA degree. 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 a BA degree give or take ___ \(\%\). d. Suppose we decreased the sample size from 500 to 100 . What effect would this have on the standard error? Recalculate the standard error to see if your prediction was correct.

a. If a rifleman's gunsight is adjusted incorrectly, he might shoot bullets consistently close to 2 feet left of the bull's-eye target. Draw a sketch of the target with the bullet holes. Does this show lack of precision or bias? b. Draw a second sketch of the target if the shots are both unbiased and precise (have little variation). The rifleman's aim is not perfect, so your sketches should show more than one bullet hole.

From Formula \(7.2\), an estimate for margin of error for a \(95 \%\) confidence interval is \(m=2 \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\) where \(\mathrm{n}\) is the required sample size and \(\hat{p}\) is the sample proportion. Since we do not know a value for \(\hat{p}\), we use a conservative estimate of \(0.50\) for \(\hat{p}\). Replace \(\hat{p}\) with \(0.50\) in the formula and simplify.
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