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Chapter 6: Problem 33

A margin of error within \(\pm 2 \%\) with \(95 \%\) confidence. An initial small sample has \(\hat{p}=0.78\).

Short Answer

Expert verified
To apply the formula and conduct the calculations, you need a sample size of approximately 2401 to achieve an error margin within 卤2 % with 95 % confidence level if the initial small sample proportion is 0.78.

Step by step solution

01

Understand the Formula

First, one needs to understand the formula for calculating minimum sample size in proportions is \(n= \frac{\(Z^2\) * \(\hat{p}(1-\hat{p})}{E^2}\), where \(Z\) is the Z-score, \(\hat{p}\) is the estimated proportion, and \(E\) is the margin of error.
02

Calculate Z-score

For a 95% confidence level, the Z-score is approximately 1.96. This value is obtained from the standard normal distribution table.
03

Substitution into the Formula

Substitute the given values into the formula. The estimated proportion \(\hat{p}\) is 0.78 and the margin of error \(E\) is 0.02. Now, use the formula \(n = \frac{(1.96)^2 * 0.78 * (1 - 0.78)}{(0.02)^2}\).
04

Solving the Formula

Calculate the equation obtained in the previous step to find the minimum sample size. If you get a non-integer value, always round up because you can't have a fraction of a sample.

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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 statistic that expresses the amount of random sampling error in the results of a survey. It represents how much the survey results will differ from the true population value.

For example, if a political poll reports that a candidate has 50% support with a margin of error of \(\pm 3\text{%}\), then the true support is likely between 47% and 53%. In the exercise, a \(\pm 2\text{%}\) margin of error signifies the researcher's willingness to accept that their survey's estimate could be off by up to 2 percentage points from the actual population proportion.
Confidence Level
A confidence level refers to the probability that the value of a parameter falls within a specified range of values, typically expressed as a percentage. Most commonly used confidence levels are 90%, 95%, or 99%.

Choosing a 95% confidence level means that if the same population is sampled numerous times and interval estimates are made on each occasion, the actual population parameter will be in the interval estimates 95% of the time. The remaining 5% represents the risk you are taking of being wrong.
Z-score
The Z-score is a measure indicating how many standard deviations an element is from the mean. In the context of sample size calculation, the Z-score represents the number of standard deviations a certain confidence level requires, corresponding to the tail(s) of a normal distribution.

For a 95% confidence level, the Z-score is 1.96, which is derived from the standard normal distribution and indicates the range within which the true population proportion is likely to be found 95% of the time.
Estimated Proportion
The estimated proportion, often denoted as \(\hat{p}\), is a statistic that estimates the fraction of the population possessing a particular attribute, based on the sample data.

In the given exercise, \(\hat{p}=0.78\) indicates that in the initial small sample, 78% of the observed subjects had the characteristic of interest. This estimated proportion is crucial in determining the variability in the population, which directly affects the sample size needed.
Sample Size Calculation
Sample size calculation is the process of determining the number of observations or replicates to include in a statistical sample. The formula used in the exercise is \(n = \frac{Z^2 * \hat{p}(1-\hat{p})}{E^2}\), incorporating the Z-score, estimated proportion, and margin of error to ensure that the sample adequately represents the population.

This formula assumes a simple random sample and a normal distribution of the population proportion. The result gives the minimum sample size needed to estimate the true population proportion with the desired confidence and precision.

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Most popular questions from this chapter

Number of Bedrooms in Houses in New York and New Jersey The dataset HomesForSale has data on houses available for sale in three Mid-Atlantic states (NY, NJ, and PA). For this exercise we are specifically interested in homes for sale in New York and New Jersey. We have information on 30 homes from each state and observe the proportion of homes with more than three bedrooms. We find that \(26.7 \%\) of homes in NY \(\left(\hat{p}_{N Y}\right)\) and \(63.3 \%\) of homes in NJ \(\left(\hat{p}_{N J}\right)\) have more then three bedrooms. (a) Is the normal distribution appropriate to model this difference? (b) Test for a difference in proportion of homes with more than three bedrooms between the two states and interpret the result.

Table 6.29 gives a sample of grades on the first two quizzes in an introductory statistics course. We are interested in testing whether the mean grade on the second quiz is significantly higher than the mean grade on the first quiz. (a) Complete the test if we assume that the grades from the first quiz come from a random sample of 10 students in the course and the grades on the second quiz come from a different separate random sample of 10 students in the course. Clearly state the conclusion. (b) Now conduct the test if we assume that the grades recorded for the first quiz and the second quiz are from the same 10 students in the same order. (So the first student got a 72 on the first quiz and a 78 on the second quiz.) (c) Why are the results so different? Which is a better way to collect the data to answer the question of whether grades are higher on the second quiz? $$ \begin{array}{llllllllll} \hline \text { First Quiz } & 72 & 95 & 56 & 87 & 80 & 98 & 74 & 85 & 77 & 62 \\\ \text { Second Quiz } & 78 & 96 & 72 & 89 & 80 & 95 & 86 & 87 & 82 & 75 \\ \hline \end{array} $$

We see in the AllCountries dataset that the percent of the population that is elderly (over 65 years old) is 17.0 in Austria and 15.9 in Denmark. Suppose we take random samples of size 200 from each of these countries and compute the difference in sample proportions \(\hat{p}_{A}-\hat{p}_{D}\), where \(\hat{p}_{A}\) represents the proportion of the sample that is elderly in Austria and \(\hat{p}_{D}\) represents the proportion of the sample that is elderly in Denmark. Find the mean and standard deviation of the differences in sample proportions.

In Exercises 6.153 to \(6.158,\) if random samples of the given sizes are drawn from populations with the given proportions: (a) Find the mean and standard error of the distribution of differences in sample proportions, \(\hat{p}_{A}-\hat{p}_{B}\) (b) If the sample sizes are large enough for the Central Limit Theorem to apply, draw a curve showing the shape of the sampling distribution. Include at least three values on the horizontal axis. Samples of size 50 from population \(A\) with proportion 0.70 and samples of size 75 from population \(B\) with proportion 0.60

Test \(H_{0}: \mu_{1}=\mu_{2}\) vs \(H_{a}: \mu_{1} \neq \mu_{2}\) using the paired difference sample results \(\bar{x}_{d}=-2.6, s_{d}=4.1\) \(n_{d}=18\)
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