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Basic Computation: \(\hat{p}\) Distribution Suppose we have a binomial experiment in which success is defined to be a particular quality or attribute that interests us. (a) Suppose \(n=100\) and \(p=0.23 .\) Can we safely approximate the \(\hat{p}\) distribution by a normal distribution? Why? Compute \(\mu_{j}\) and \(\sigma_{\tilde{p}}\) (b) Suppose \(n=20\) and \(p=0.23 .\) Can we safely approximate the \(\hat{p}\) distribution by a normal distribution? Why or why not?

Short Answer

Expert verified
(a) Yes, can approximate; \(\mu_{\hat{p}} = 0.23\) and \(\sigma_{\hat{p}} = 0.042\). (b) No, cannot approximate.

Step by step solution

01

Check Normal Approximation Condition for Part (a)

To use the normal approximation of a binomial distribution, we check if both \(np\) and \(n(1-p)\) are greater than or equal to 5. For \(n = 100\) and \(p = 0.23\), calculate \(np\) and \(n(1-p)\):\[ np = 100 \times 0.23 = 23 \]\[ n(1-p) = 100 \times (1-0.23) = 77 \]Since both values are greater than 5, we can safely approximate the distribution by a normal distribution.
02

Compute Mean and Standard Deviation for Part (a)

The mean \(\mu_{\hat{p}}\) of the sample proportion \(\hat{p}\) is simply \(p\), and the standard deviation \(\sigma_{\hat{p}}\) is given by:\[ \sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.23 \times (1-0.23)}{100}} = \sqrt{\frac{0.1771}{100}} = 0.042 \]Thus, \(\mu_{\hat{p}} = 0.23\) and \(\sigma_{\hat{p}} = 0.042\).
03

Check Normal Approximation Condition for Part (b)

Again, to use the normal approximation, we calculate \(np\) and \(n(1-p)\) for \(n = 20\) and \(p = 0.23\):\[ np = 20 \times 0.23 = 4.6 \]\[ n(1-p) = 20 \times (1-0.23) = 15.4 \]Here, \(np\) is less than 5, thus we cannot safely approximate the distribution by a normal distribution for \(n = 20\).

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Key Concepts

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

Binomial Distribution
The binomial distribution is the foundation for understanding certain types of random experiments. It models situations where there are two possible outcomes—success and failure. Consider a coin flip, where landing on heads is a success and tails is a failure. In a binomial experiment, the number of trials is fixed, and each trial is independent. Additionally, the probability of success remains constant across trials.

To identify a situation as a binomial distribution, ask yourself these questions:
  • Is there a fixed number of trials?
  • Are each of the trials independent?
  • Are there exactly two outcomes per trial?
  • Is the probability of success the same for each trial?
If the answer is yes to all, you are likely dealing with a binomial distribution. Understanding this allows one to compute probabilities that a given number of successes will occur in the experiment.
Normal Approximation
Normal approximation is a way to simplify the analysis of a binomial distribution by using the normal distribution. This is especially useful when working with a large number of trials, where calculations with the binomial formula can become cumbersome.

For normal approximation to be valid, two conditions should be satisfied:
  • \( np \geq 5 \): The expected number of successes should be at least 5.
  • \( n(1-p) \geq 5 \): The expected number of failures should also be at least 5.
When these conditions are met, the binomial distribution of sample proportions \( \hat{p} \) can be considered bell-shaped and symmetric, making it similar to a normal distribution. This approximation allows us to use easier-to-understand normal distribution techniques.
Sample Proportion
The sample proportion \( \hat{p} \) represents the fraction of successes found in a sample from a binomial distribution. If you surveyed 100 people and 23 of them preferred chocolate over vanilla ice cream, your sample proportion \( \hat{p} \) would be 0.23.

Sample proportions allow statisticians to draw inferences about the population proportion without surveying everyone. However, remember that the approximation works best when you have a large sample size. \( \hat{p} \) is a randomized variable because its value can vary depending on the sample taken, so it can be analyzed like a distribution itself. This makes it essential to understand its behavior, especially regarding its mean and standard deviation.
Mean and Standard Deviation
The mean and standard deviation for a sample proportion \( \hat{p} \) are essential statistics to understand its distribution.

The mean of \( \hat{p} \) is simply the population proportion \( p \) itself, meaning if you keep taking larger and larger samples, they should average out to the true proportion of successes:

  • \( \mu_{\hat{p}} = p \)
The standard deviation \( \sigma_{\hat{p}} \) provides insight into the variability of the sample proportion across different samples and is expressed by:

  • \( \sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}} \)
This formula shows that as the sample size \( n \) increases, the standard deviation decreases, indicating that your sample proportions will be more tightly clustered around the mean. Understanding this helps make an informed decision on the reliability of the approximation and statistical inference.

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Most popular questions from this chapter

The taxi and takeoff time for commercial jets is a random variable \(x\) with a mean of 8.5 minutes and a standard deviation of 2.5 minutes. Assume that the distribution of taxi and takeoff times is approximately normal. You may assume that the jets are lined up on a runway so that one taxies and takes off immediately after another, and that they take off one at a time on a given runway. What is the probability that for 36 jets on a given runway, total taxi and takeoff time will be (a) less than 320 minutes? (b) more than 275 minutes? (c) between 275 and 320 minutes? Hint: See Problem 21

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