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Just as the difference between two sample means is normally distributed for large samples, so is the difference between two sample proportions. That is, if \(Y_{1}\) and \(Y_{2}\) are independent binomial random variables with parameters \(\left(n_{1}, p_{1}\right)\) and \(\left(n_{2}, p_{2}\right),\) respectively, then \(\left(Y_{1} / n_{1}\right)-\left(Y_{2} / n_{2}\right)\) is approximately normally distributed for large values of \(n_{1}\) and \(n_{2}\) a. Find \(E\left(\frac{Y_{1}}{n_{1}}-\frac{Y_{2}}{n_{2}}\right)\) b. Find \(V\left(\frac{Y_{1}}{n_{1}}-\frac{Y_{2}}{n_{2}}\right)\)

Short Answer

Expert verified
a. \( p_1 - p_2 \); b. \( \frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2} \)."

Step by step solution

01

Understand the problem

We need to find the expected value and the variance of the difference between two sample proportions \( E\left(\frac{Y_1}{n_1} - \frac{Y_2}{n_2}\right) \) and \( V\left(\frac{Y_1}{n_1} - \frac{Y_2}{n_2}\right) \).
02

Define the random variables

Let \( Y_1 \) and \( Y_2 \) be binomial random variables with parameters \((n_1, p_1)\) and \((n_2, p_2)\) respectively. Hence, \( \frac{Y_1}{n_1} \) and \( \frac{Y_2}{n_2} \) are the sample proportions.
03

Calculate Expected Value of Proportions

For a binomial random variable \( Y_i \), the expected value \( E(Y_i) = n_i p_i \). Therefore, \( E\left(\frac{Y_1}{n_1}\right) = p_1 \) and \( E\left(\frac{Y_2}{n_2}\right) = p_2 \).
04

Use Linearity of Expectation

By the linearity of expectation, the expected value of the difference is: \( E\left(\frac{Y_1}{n_1} - \frac{Y_2}{n_2}\right) = E\left(\frac{Y_1}{n_1}\right) - E\left(\frac{Y_2}{n_2}\right) = p_1 - p_2 \).
05

Calculate Variance of Proportions

For a binomial random variable \( Y_i \), the variance \( V(Y_i) = n_i p_i (1 - p_i) \). Hence, \( V\left(\frac{Y_1}{n_1}\right) = \frac{p_1(1-p_1)}{n_1} \) and \( V\left(\frac{Y_2}{n_2}\right) = \frac{p_2(1-p_2)}{n_2} \).
06

Use Independence to Find Variance

Since \( Y_1 \) and \( Y_2 \) are independent, the variance of their difference is: \( V\left(\frac{Y_1}{n_1} - \frac{Y_2}{n_2}\right) = V\left(\frac{Y_1}{n_1}\right) + V\left(\frac{Y_2}{n_2}\right) = \frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2} \).

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Key Concepts

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

Binomial Distribution
In statistics, the binomial distribution is a discrete probability distribution. It arises from scenarios where an experiment is conducted multiple times, each with two possible outcomes: success or failure.
These experiments are known as Bernoulli trials. The binomial distribution depends on two parameters: the number of trials (\(n\)) and the probability of success of each trial (\(p\)).
  • Properties of Binomial Distribution: It is represented as \( B(n, p) \).
  • The mean, or expected value, is calculated as \( E(Y) = np \).
  • The variance is calculated as \( V(Y) = np(1-p) \).
Binomial distribution is a cornerstone in statistical inference because it helps understand phenomena with binary outcomes, such as flipping a coin or labeling outcomes as pass/fail.
In the context of the original exercise, binomial random variables \(Y_1\) and \(Y_2\) with parameters \( (n_1, p_1) \) and \( (n_2, p_2) \) respectively, allow us to analyze the sample proportions.
Sample Proportions
Sample proportions are a way of estimating population parameters. They are derived by dividing the number of successful events by the total number of trials.
For example, if you interviewed 100 people and 40 of them own a car, then the sample proportion \( \frac{40}{100} \) or 0.4 represents those who own a car.
  • Calculating Sample Proportions: For binomial random variables, sample proportions can be expressed as: \( \frac{Y_1}{n_1} \) and \( \frac{Y_2}{n_2} \) for two samples
  • They provide estimates for the actual population proportions \( p_1 \) and \( p_2 \), where \( Y_1 \) is the number of successes in the first sample and \( Y_2 \) in the second.
  • Importance in Statistical Inference: When comparing two proportions, it is crucial to understand the expected value \( p_1 - p_2 \), showing the difference between the two. This can indicate if a significant difference exists between populations.
Understanding sample proportions allows statisticians to make informed decisions about populations based on sample data.
Variance Calculation
Variance is a measure of how much data points differ from the mean in a dataset. It is particularly handy when assessing how spread out data is. For binomial variables, variance quantifies uncertainty in predictions based on samples.
Calculating variance for sample proportions involves a few steps, leveraging properties of binomial distribution:
  • Variance of Binomial Random Variable: For a binomial random variable \(Y_i\), variance is given by \( V(Y_i) = n_i p_i (1-p_i) \).
  • Variance of Sample Proportions: To find variance in sample proportions, use: \( V\left ( \frac{Y_1}{n_1} \right ) = \frac{p_1(1-p_1)}{n_1} \) and \( V\left ( \frac{Y_2}{n_2} \right ) = \frac{p_2(1-p_2)}{n_2} \) .
  • Independence in Variance Calculation: If the samples are independent, total variance is the sum of variances of proportions: \( V\left( \frac{Y_1}{n_1} - \frac{Y_2}{n_2} \right) = \frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2} \).
Variance is key in understanding the reliability of inferences made from sample data as it provides insights into data variability and prediction confidence.

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Most popular questions from this chapter

Suppose that \(T\) is a \(t\) -distributed random variable. a. If \(T\) has 5 df, use Table 5 , Appendix 3 , to find \(t_{10}\), the value such that \(P\left(T>t_{10}\right)=.10\). Find \(t_{10}\) using the applet Student's \(t\) Probabilities and Quantiles. b. Refer to part (a). What quantile does \(t_{.10}\) correspond to? Which percentile? c. Use the applet Student's t Probabilities and Quantiles to find the value of \(t_{.10}\) for \(t\) distributions with 30, 60, and 120 df. d. When \(Z\) has a standard normal distribution, \(P(Z>1.282)=.10\) and \(z_{.10}=1.282 .\) What property of the \(t\) distribution (when compared to the standard normal distribution) explains the fact that all of the values obtained in part (c) are larger than \(z_{10}=1.282 ?\) e. What do you observe about the relative sizes of the values of \(t_{.10}\) for \(t\) distributions with 30,60 and \(120 \mathrm{df} ?\) Guess what \(t_{.10}\) "converges to" as the number of degrees of freedom gets large. [Hint: Look at the row labeled \(\infty \text { in Table } 5 \text { , Appendix } 3 .\)]

Many bulk products-such as iron ore, coal, and raw sugar- -are sampled for quality by a method that requires many small samples to be taken periodically as the material is moving along a conveyor belt. The small samples are then combined and mixed to form one composite sample. Let \(Y_{i}\) denote the volume of the \(i\) th small sample from a particular lot and suppose that \(Y_{1}, Y_{2}, \ldots, Y_{n}\) constitute a random sample, with each \(Y_{i}\) value having mean \(\mu\) (in cubic inches) and variance \(\sigma^{2}\). The average volume \(\mu\) of the samples can be set by adjusting the size of the sampling device. Suppose that the variance \(\sigma^{2}\) of the volumes of the samples is known to be approximately 4. The total volume of the composite sample must exceed 200 cubic inches with probability approximately .95 when \(n=50\) small samples are selected. Determine a setting for \(\mu\) that will allow the sampling requirements to be satisfied.

In 2004 Florida was hit by four major hurricanes. In 2005 a survey indicated that, in \(2004,48 \%\) of the households in Florida had no plans for escaping an approaching hurricane. Suppose that a recent random sample of 50 households was selected in Gainesville and that those in 29 of the households indicated that their household had a hurricane escape plan. a. If the 2004 state percentages still apply to recent Gainesville households, use the Normal Approximation to Binomial Distribution applet to find the exact and approximate values of the probability that 29 or more of the households sampled have a hurricane escape plan. b. Refer to part (a). Is the normal approximation close to the exact binomial probability? Explain why.

The times to process orders at the service counter of a pharmacy are exponentially distributed with mean 10 minutes. If 100 customers visit the counter in a 2 -day period, what is the probability that at least half of them need to wait more than 10 minutes?

An anthropologist wishes to estimate the average height of men for a certain race of people. If the population standard deviation is assumed to be 2.5 inches and if she randomly samples 100 men, find the probability that the difference between the sample mean and the true population mean will not exceed .5 inch.

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