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The sample size needed to estimate the difference between two population proportions to within a margin of error \(E\) with a confidence level of \(1-\alpha\) can be found by using the following expression: $$ E=z_{\alpha / 2} \sqrt{\frac{p_{1} q_{1}}{n_{1}}+\frac{p_{2} q_{2}}{n_{2}}} $$ Replace \(n_{1}\) and \(n_{2}\) by \(n\) in the preceding formula (assuming that both samples have the same size) and replace each of \(p_{1}, q_{1}, p_{2}\), and \(q_{2}\) by \(0.5\) (because their values are not known). Solving for \(n\) results in this expression: $$ n=\frac{z_{\alpha / 2}^{2}}{2 E^{2}} $$ Use this expression to find the size of each sample if you want to estimate the difference between the proportions of men and women who own smartphones. Assume that you want \(95 \%\) confidence that your error is no more than \(0.03\).

Step by step solution

01

- Identify the given values

Extract the values provided in the problem statement: 1. Confidence level: 95% 2. Margin of error (E): 0.03

02

- Find the z-value for the given confidence level

Since the confidence level is 95%, find the z-value for \(\frac{\text{confidence level}}{2}\). This corresponds to \ z_{\frac{0.05}{2}} \ or \ z_{0.025}. \ The z-value can be found using a z-table or standard normal distribution and is approximately 1.96.

03

- Use the given expression to calculate the sample size

Insert the z-value and the margin of error into the given formula: \( n = \frac{z_{\frac{\text{confidence level}}{2}}^2}{2E^2} \). So, substituting the values: \( n = \frac{1.96^2}{2 \times 0.03^2} \).

04

- Perform the calculations

Calculate the numerator: \( 1.96^2 = 3.8416 \)Calculate the denominator: \( 2 \times 0.03^2 = 2 \times 0.0009 = 0.0018 \)Finally, divide the numerator by the denominator to find the sample size n: \( n = \frac{3.8416}{0.0018} \)

05

- Determine the final sample size

Perform the final calculation: \( n = 2134.2222 \).Since the sample size must be a whole number, round up to the nearest whole number: \( n \) = 2135.

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Key Concepts

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

confidence level

The confidence level represents how sure you are about your estimate being close to the true population parameter. For instance, if you have a 95% confidence level, it means you are 95% certain that the true value lies within the specified margin of error. This concept is often used in conjunction with sample size estimation to determine how many samples you need to estimate a population parameter accurately.
The confidence level is closely related to the z-value, which you can find in a z-table or using statistical software. It translates your confidence percentage into a critical value, showing how many standard deviations away from the mean your results should be.

margin of error

The margin of error tells you how much error you are willing to accept in your estimation. For example, if the margin of error is 0.03, this means that the true population parameter could be off by up to 0.03 in either direction. It's a measure of the range of uncertainty in your estimate.
The margin of error is crucial when calculating the sample size because it directly influences how precise your estimate will be. The smaller the margin of error, the more accurate your estimate, but this typically requires a larger sample size.

population proportions

Population proportions represent the fraction of the population that possesses a particular characteristic. When dealing with categories like the proportion of men and women who own smartphones, these proportions provide valuable insights.
If you don't have prior information about population proportions, a common approach is to assume that each proportion is 0.5. This is because using 0.5 generally provides the largest possible sample size, ensuring your estimate is conservative and reliable.

z-value

The z-value, also known as the z-score, is a value from the standard normal distribution, which is used to determine the critical value in hypothesis testing and confidence intervals. For example, a 95% confidence level corresponds to a z-value of approximately 1.96.
This value is essential in the formula for sample size estimation, as it helps determine how many standard deviations away from the mean you should be to achieve the desired confidence level.

standard normal distribution

The standard normal distribution is a special type of normal distribution where the mean is 0 and the standard deviation is 1. This distribution is widely used in statistics because it allows easy computation of probabilities and z-values.
When you look up z-values in a z-table, you're using the standard normal distribution. This distribution helps convert real-world data to standardized data, making it easier to draw meaningful conclusions and estimate sample sizes.