/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 64 The article "Should Canada Allow... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The article "Should Canada Allow Direct-to-Consumer Advertising of Prescription Drugs?" (Canadian Family Physician [2009]: \(130-131\) ) calls for the legalization of advertising of prescription drugs in Canada. Suppose you wanted to conduct a survey to estimate the proportion of Canadians who would allow this type of advertising. How large a random sample would be required to estimate this proportion with a margin of error of \(0.02 ?\)

Short Answer

Expert verified
A random sample size of \(2401\) is required to estimate the proportion of Canadians who would allow direct-to-consumer advertising of prescription drugs with a margin of error of \(0.02\), assuming a \(95\%\) confidence level.

Step by step solution

01

Identify the given information

In this problem, we're given the following information: - Margin of error (E): 0.02 Now, we need to recall the formula for finding the sample size in a proportion estimation problem. Remember, we want to have a sample size large enough so that the margin of error is 0.02.
02

Recall the sample size formula

In determining the sample size (n) for estimating a proportion, we'll use the following formula: \[ n = \frac{Z^2 * p * (1 - p)}{E^2} \] where: - n: sample size - Z: Z-score for desired confidence level (we'll use a 95% confidence level, which corresponds to a Z-score of 1.96) - p: estimated proportion (since we don't have an estimate, we'll use the most conservative estimate of 0.5) - E: margin of error (0.02)
03

Calculate the sample size

Using the formula from step 2, substitute the given values and solve for n: \[ n = \frac{1.96^2 * 0.5 * (1 - 0.5)}{0.02^2} \] \[ n = \frac{(3.8416) * (0.5) * (0.5)}{0.0004} \] \[ n = 2401 \]
04

Interpret the result

We found that the required random sample size to estimate the proportion of Canadians who would allow direct-to-consumer advertising of prescription drugs with a margin of error of 0.02 is 2401. Keep in mind that this sample size assumes a 95% confidence level and a conservative estimate for the true proportion. The actual required sample size may be smaller if the true proportion is different from 0.5.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Margin of Error
The margin of error is a statistical term that represents the amount of error that can be tolerated in the sample estimate. It specifies the range within which the true population parameter is expected to lie with a certain degree of confidence. For example, when surveying individuals about their opinion on a topic, the margin of error indicates how close the sample's proportion is likely to reflect the actual proportion in the entire population.

When the margin of error is small, as in the case of the Canadian survey on advertising prescription drugs mentioned in the exercise, the result is a more precise estimate but generally requires a larger sample size. Conversely, a larger margin of error means less precision but can be achieved with a smaller sample size. In our example, a margin of error of 0.02 allows for a very accurate estimate of the population's sentiment, ensuring that the true proportion should fall within 2 percentage points of the sample proportion.
Confidence Level
The confidence level is a measure of how certain we are that the population parameter lies within the margin of error. It's expressed as a percentage and reflects the frequency with which the confidence interval will contain the true population parameter over many samples. A higher confidence level means we are more certain that the true parameter is within the given range.

A 95% confidence level is commonly used, stating that if we were to take 100 random samples from the population, we would expect the true population proportion to be within the margin of error in 95 of those samples. This high level of confidence is often sought in research, as in the case of determining Canadian support for prescription drug advertising, to ensure the survey results are reliable and can be trusted to make informed decisions.
Proportion Estimation
Proportion estimation involves estimating the percentage of a certain characteristic within a population based on a sample. In the exercise, the survey aims to estimate the proportion of Canadians who favor the legalization of direct-to-consumer advertising for prescription drugs.

To achieve a good estimate, it is vital to use a representative sample of the population and an appropriate sample size. The most conservative estimate of the proportion (when no prior data is available) is 0.5, meaning we do not lean towards assuming the majority for or against the issue. This conservative approach ensures that the maximum sample size is calculated, which would provide the most accuracy in the estimation under the given margin of error and confidence level.
Z-score
A Z-score is a statistical measurement that describes a value's relationship to the mean of a group of values, measured in terms of standard deviations from the mean. In the context of sample size determination, the Z-score is used to find the critical value corresponding to the chosen confidence level.

In our survey example, the 95% confidence level corresponds to a Z-score of approximately 1.96. This value reflects the number of standard deviations a datum must be from the mean to be within the top 2.5% of a normal distribution on either side. When calculating the sample size needed, the Z-score helps determine how far the sample proportion could deviate from the population proportion and still achieve the desired confidence level. A higher Z-score is associated with a higher confidence level and, consequently, a larger required sample size for the same margin of error.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The article "Career Expert Provides DOs and DON'Ts for Job Seekers on Social Networking" (CareerBuilder.com, August 19,2009 ) included data from a survey of 2667 hiring managers and human resource professionals. The article noted that more employers are now using social networks to screen job applicants. Of the 2667 people who participated in the survey, 1200 indicated that they use social networking sites such as Facebook, MySpace, and LinkedIn to research job applicants. Assume that the sample is representative of hiring managers and human resource professionals. Answer the four key questions (QSTN) to confirm that the suggested method in this situation is a confidence interval for a population proportion.

A random sample will be selected from the population of all adult residents of a particular city. The sample proportion \(\hat{p}\) will be used to estimate \(p,\) the proportion of all adult residents who are registered to vote. For which of the following situations will the estimate tend to be closest to the actual value of \(p ?\) I. \(\quad n=1000, p=0.5\) II. \(\quad n=200, p=0.6\) III. \(n=100, p=0.7\)

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?

A researcher wants to estimate the proportion of students enrolled at a university who are registered to vote. Would the standard error of the sample proportion \(\hat{p}\) be larger if the actual population proportion was \(p=0.4\) or \(p=0.8 ?\)

A researcher wants to estimate the proportion of city residents who favor spending city funds to promote tourism. Would the standard error of the sample proportion \(\hat{p}\) be smaller for random samples of size \(n=100\) or random samples of size \(n=200 ?\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.