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Why do we use \(1 / 4\) in place of \(p(1-p)\) in formula (22) for sample size when the probability of success \(p\) is unknown? (a) Show that \(p(1-p)=1 / 4-(p-1 / 2)^{2}\). (b) Why is \(p(1-p)\) never greater than \(1 / 4 ?\)

Short Answer

Expert verified
The value of \( p(1-p) \) is never greater than \( 1/4 \) because it is maximized at \( p = 1/2 \), where \( p(1-p) = 1/4 \).

Step by step solution

01

Expand the squared term

To show that \( p(1-p) = \frac{1}{4} - (p - \frac{1}{2})^2 \), we first expand the squared term. The expression \((p - \frac{1}{2})^2\) expands to \( p^2 - p + \frac{1}{4} \).
02

Rearrange the expression

Now rewrite \( p(1-p) \) as \( p - p^2 \). Substitute the expanded square into the right-hand side of the equation: \( \frac{1}{4} - (p^2 - p + \frac{1}{4}) \).
03

Simplify the expression

Distribute the negative sign: \( \frac{1}{4} - p^2 + p - \frac{1}{4} \). This simplifies to \( p - p^2 \), which equals \( p(1-p) \). This confirms that \( p(1-p) = \frac{1}{4} - (p - \frac{1}{2})^2 \).
04

Analyze the maximum of p(1-p)

The expression \( (p - \frac{1}{2})^2 \) is a square term, meaning it is always non-negative. Hence, \( \frac{1}{4} - (p - \frac{1}{2})^2 \) is maximized when \( (p - \frac{1}{2})^2 = 0 \), that is, when \( p = \frac{1}{2} \).
05

Conclusion on the maximum value

Since the maximum value is achieved when \( p = \frac{1}{2} \) and the expression becomes \( \frac{1}{4} - 0 = \frac{1}{4} \), the value of \( p(1-p) \) can never be greater than \( \frac{1}{4} \). This is why \( \frac{1}{4} \) is used as a conservative estimate when \( p \) is unknown.

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Key Concepts

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

Probability of Success
The probability of success, often denoted as \( p \), is an essential concept when determining sample sizes in statistics. It refers to the likelihood of a certain event occurring. When \( p \) is unknown, estimating it can be challenging. However, understanding how probabilities work is crucial in scenarios where such estimates are needed. For sample size determination, the behavior of \( p(1-p) \) helps us understand the variability of the sample proportion.When \( p \) is unknown, we use \( \frac{1}{4} \) instead of \( p(1-p) \) in formula calculations. This is because this fraction represents the maximum value \( p(1-p) \) can achieve, making it a safe and conservative estimate. By using this maximum value, we ensure that calculations do not underestimate the sample size needed for accurate results.
Quadratic Expressions
Quadratic expressions are algebraic expressions that involve terms where the variable is squared. In the context of the given exercise, the expression \( (p - \frac{1}{2})^2 \) plays a significant role.Here's how it works:
  • The original expression \( p(1-p) \) can be rearranged to show its similarity to a quadratic expression.
  • By expanding \( (p - \frac{1}{2})^2 \), we arrive at \( p^2 - p + \frac{1}{4} \), which reveals how these terms can be reorganized to link them back to \( p(1-p) \).
This connection is not just a simple algebraic identity; it reveals deeper insights into how the probability of success fluctuates with variation in \( p \). Understanding such expressions is vital for grasping the underlying statistical principles related to probability.
Maximum Value Analysis
Maximum value analysis comes into play when determining constraints or limits within mathematical expressions. In the case of \( p(1-p) \), determining its maximum value is key to sample size estimation.The expression \( (p - \frac{1}{2})^2 \) is always non-negative, meaning it can't go below zero. Thus, to maximize \( \frac{1}{4} - (p - \frac{1}{2})^2 \), \( (p - \frac{1}{2})^2 \) must equal zero. This occurs precisely when \( p = \frac{1}{2} \).At this point, \( \frac{1}{4} - 0 = \frac{1}{4} \), which is identified as the upper limit for \( p(1-p) \). This conservative approach eliminates the risk of underestimating how large a sample should be when \( p \) is unknown, thus ensuring accurate and reliable results based on maximum variability.

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Most popular questions from this chapter

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