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Determine a method for constructing a confidence interval for \(p_{1}-p_{2}\), the difference of two population proportions.

Short Answer

Expert verified
To construct a confidence interval for the difference between two population proportions, \(p_1 - p_2\), follow these steps: 1. Calculate the point estimate \(\hat{P} = \hat{p}_1 - \hat{p}_2\). 2. Compute the standard error \(\text{SE}(\hat{P}) = \sqrt{\frac{\hat{p}_1(1- \hat{p}_1)}{n_1} + \frac{\hat{p}_2(1- \hat{p}_2)}{n_2}}\). 3. Determine the appropriate z-value based on the desired confidence level. 4. Calculate the margin of error: \(\text{Margin of Error} = Z \times \text{SE}(\hat{P})\). 5. Construct the confidence interval: \(\text{Confidence Interval} = (\hat{P} - \text{Margin of Error}, \hat{P} + \text{Margin of Error})\).

Step by step solution

01

Find the point estimate of the difference between the proportions

The point estimate of the difference between the two proportions can be calculated using the sample proportions, denoted as \(\hat{p}_1\) and \(\hat{p}_2\). The point estimate for the difference is simply the difference between these two sample proportions: \[\hat{P} = \hat{p}_1 - \hat{p}_2\]
02

Calculate the standard error

Next, we compute the standard error of this difference. The standard error is given by: \[\text{SE}(\hat{P}) = \sqrt{\frac{\hat{p}_1(1- \hat{p}_1)}{n_1} + \frac{\hat{p}_2(1- \hat{p}_2)}{n_2}}\] Here, \(n_1\) and \(n_2\) are the sample sizes for the two populations.
03

Determine the appropriate Z-value

To construct the confidence interval, we need to find the appropriate z-value based on the desired confidence level. The most common confidence levels are 90%, 95%, and 99%. The z-value corresponding to these confidence levels are 1.645, 1.96, and 2.576 respectively.
04

Compute the margin of error

Now, we will compute the margin of error using the standard error and the z-value. The margin of error is given by: \[\text{Margin of Error} = Z \times \text{SE}(\hat{P})\]
05

Construct the confidence interval

Finally, we construct the confidence interval for the difference between the two population proportions using the point estimate and the margin of error. The confidence interval is given by: \[\text{Confidence Interval} = (\hat{P} - \text{Margin of Error}, \hat{P} + \text{Margin of Error})\] To construct a confidence interval for the difference between two population proportions, follow these steps: 1. Calculate the point estimate of the difference between the proportions, which is the difference between the sample proportions, \(\hat{p}_1\) and \(\hat{p}_2\). 2. Compute the standard error based on the sample proportions and the sample sizes. 3. Determine the appropriate z-value based on the desired confidence level. 4. Calculate the margin of error using the standard error and the z-value. 5. Construct the confidence interval using the point estimate and the margin of error.

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Key Concepts

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

Confidence Interval
When you're comparing two population proportions, such as the percentage of people favoring two different products, you often want to determine how confident you can be about the difference between these two proportions. This is where the concept of a confidence interval comes in. A confidence interval gives you a range within which the true difference between the proportions is likely to lie, based on your sample data.

To calculate a confidence interval for the difference in population proportions, follow these key steps:
  • First, establish your point estimate, which is simply the difference between the sample proportions (more on that later).
  • Next, figure out the standard error, which helps measure the variability of your estimate.
  • Finally, multiply the standard error by a z-value that corresponds to your desired confidence level (like 1.96 for 95% confidence) to find your margin of error.

By adding and subtracting this margin of error from your point estimate, you get your confidence interval. This interval estimates the range where the true difference in proportions lies, usually with a common confidence level like 95%.
Standard Error
The standard error is crucial for understanding how much your point estimate might vary if you took different samples from the populations. It's all about accounting for sample variability. For the difference between two population proportions, the formula for the standard error is:

\[\text{SE}(\hat{P}) = \sqrt{\frac{\hat{p}_1(1- \hat{p}_1)}{n_1} + \frac{\hat{p}_2(1- \hat{p}_2)}{n_2}}\]

In this formula:
  • \(\hat{p}_1\) and \(\hat{p}_2\) represent the sample proportions.
  • \(n_1\) and \(n_2\) are the sizes of these samples.
The square roots and fractions help standardize variability, even if the two sample sizes are different. A larger standard error implies more uncertainty around your point estimate, which will affect the width of your confidence interval.

Always remember that the standard error indicates how much your sample's point estimate might "move around" if you repeated your sampling process.
Point Estimate
A point estimate offers a single best guess or approximation of the true value you're interested in. In the case of comparing two population proportions, it represents the difference between the proportions from your samples. Mathematically, it is noted as:

\[\hat{P} = \hat{p}_1 - \hat{p}_2\]

Here:
  • \(\hat{p}_1\) is the sample proportion from the first group.
  • \(\hat{p}_2\) is the sample proportion from the second group.
This subtraction gives us a straightforward estimate of how much more one proportion is compared to the other, based on the samples you have collected.

However, keep in mind that this point estimate doesn't tell the whole story by itself. Since it can vary from sample to sample, we use it with the confidence interval to understand the reliability of the estimate.

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Most popular questions from this chapter

In one income group, \(45 \%\) of a random sample of people express approval of a product. In another income group, \(55 \%\) of a random sample of people express approval. The standard errors for these percentages are \(.04\) and \(.03\) respectively. Test at the \(10 \%\) level of significance the hypothesis that the percentage of people in the second income group expressing approval of the product exceeds that for the first income group.

Let \(\mathrm{X}\) have the probability distribution defined by $$ \begin{array}{lcc} \mathrm{f}(\mathrm{x})=1-\mathrm{e}^{-\mathrm{x}} & \text { for } & \mathrm{x} \geq 0 \\ \text { and }=0 & \text { for } & \mathrm{x}<0 . \end{array} $$ Let \(\mathrm{Y}=\sqrt{\mathrm{X}}\) be a new random variable. Find \(\mathrm{G}(\mathrm{y})\), the distribution function of \(\mathrm{Y}\), using the cumulative distribution function technique.

In investigating several complaints concerning the weight of the "NET WT. 12 OZ." jar of a local brand of peanut butter, the Better Business Bureau selected a sample of 36 jars. The sample showed an average net weight of \(11.92\) ounces and a standard deviation of \(.3\) ounce. Using a \(.01\) level of significance, what would the Bureau conclude about the operation of the local firm?

For a large sample, the distribution of \(\underline{X}\) is always approximately normal. Find the probability that a random sample mean lies within a) one standard error of the mean. b) two standard errors.

Let \(\mathrm{X}\) and \(\mathrm{Y}\) be jointly distributed with density function $$ \begin{array}{rlrl} \mathrm{f}(\mathrm{x}, \mathrm{y})= & 1 & 0<\mathrm{x}<1 \\ & & 0<\mathrm{y}<1 \\ & 0 & & \text { otherwise. } \end{array} $$ $$ \text { Find } \quad F(\lambda \mid X>Y)=\operatorname{Pr}(X \leq \lambda \mid X>Y) \text { . } $$

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