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The proportion of residents in Phoenix favoring the building of toll roads to complete the freeway system is believed to be \(p=0.3\). If a random sample of 10 residents shows that 1 or fewer favor this proposal, we will conclude that \(p<0.3\). (a) Find the probability of type I error if the true proportion is \(p=0.3\) (b) Find the probability of committing a type II error with this procedure if \(p=0.2\) (c) What is the power of this procedure if the true proportion is \(p=0.2 ?\)

Short Answer

Expert verified
(a) Type I error: 0.6172; (b) Type II error: 0.6778; (c) Power: 0.3222.

Step by step solution

01

Understanding Type I Error

A Type I error occurs when we reject the null hypothesis when it is actually true. In this problem, it is when we conclude that \( p < 0.3 \) while the true proportion is actually \( p = 0.3 \). We need to calculate the probability of observing 1 or fewer residents favoring the proposal under this condition.
02

Calculate Probability for Type I Error

The probability of 1 or fewer residents favoring the proposal in a sample of 10 when \( p = 0.3 \) follows a binomial distribution: \( X \sim B(n=10, p=0.3) \). We need to find \( P(X \leq 1) \). \[P(X \leq 1) = P(X = 0) + P(X = 1)\]Calculate \( P(X = 0) = \binom{10}{0} (0.3)^0 (0.7)^{10} = 0.7^{10} \)Calculate \( P(X = 1) = \binom{10}{1} (0.3)^1 (0.7)^9 = 10 \times 0.3 \times 0.7^9 \)Add these probabilities to find \( P(X \leq 1) \).
03

Understanding Type II Error

A Type II error occurs when we fail to reject the null hypothesis even though the alternative hypothesis is true. Here, it means not rejecting \( p = 0.3 \) when the true proportion is actually \( p = 0.2 \). We must find \( P(X > 1) \) when \( p = 0.2 \).
04

Calculate Probability for Type II Error

Again, use a binomial distribution: \( X \sim B(n=10, p=0.2) \). We need \( P(X > 1) = 1 - P(X \leq 1) \).First, calculate \( P(X \leq 1) = P(X = 0) + P(X = 1) \).\[P(X = 0) = \binom{10}{0} (0.2)^0 (0.8)^{10} = 0.8^{10}\]\[P(X = 1) = \binom{10}{1} (0.2)^1 (0.8)^9 = 10 \times 0.2 \times 0.8^9\]Compute \( 1 - P(X \leq 1) \).
05

Understanding Power of the Test

The power of a test is the probability of correctly rejecting the null hypothesis when the alternative hypothesis is true. For this problem, it is the probability of rejecting \( p = 0.3 \) when \( p = 0.2 \). It is calculated as \( 1 \) minus the probability of a Type II error.
06

Calculate Power of the Test

Subtract the probability of a Type II error from 1.\[\text{Power} = 1 - P(X > 1 \text{ when } p = 0.2) = 1 - \text{Probability of Type II Error}\]This results in the power of the test, ensuring that we understand the likelihood of correctly concluding that \( p < 0.3 \) when \( p = 0.2 \).

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Key Concepts

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

Type I Error
The concept of a Type I error is crucial when performing hypothesis testing. It occurs when you incorrectly reject the null hypothesis. In simpler terms, it means concluding that there is an effect or a difference when, in fact, there isn't. In the context of our exercise, a Type I error would happen if we decide that less than 30% of residents favor the toll road proposal, although exactly 30% do favor it. To determine the chance of making a Type I error, you would calculate the probability of getting 1 or fewer residents out of a sample of 10 favoring the proposal if the true proportion is indeed 0.3. This involves the use of a binomial distribution formula, which helps in finding how often this event occurs by chance.
Type II Error
A Type II error is slightly different and occurs when you fail to reject the null hypothesis even when the alternative hypothesis should actually be accepted. You can think of it as a missed opportunity to detect a real effect or difference. In our specific problem, this error would occur if we don't reject the hypothesis that 30% of residents support the toll roads when actually only 20% do. Calculating the probability of a Type II error involves looking at scenarios where more than 1 resident in the sample supports the proposal when, in reality, the true supporter proportion is less, say 20% in this example. This also uses the binomial distribution, enabling an assessment of how likely it is to overlook the real proportion due to sampling variability.
Binomial Distribution
The binomial distribution is an important tool used in probability and statistics, especially when dealing with yes/no results in a fixed number of trials. It is crucial for hypothesis testing involving proportions, like in our toll roads exercise. Here, each resident either favors or doesn't favor building toll roads, making it a classic binomial scenario. For a fixed number, n (10 residents), we assess how likely various outcomes are, given a certain probability, p (e.g., 0.3 or 0.2). The distribution allows us to compute the probability of X residents favoring the proposal. It is important as it provides the foundation for calculating both Type I and Type II errors, as these errors rely on assessing unlikely outcomes within this distribution.
Power of a Test
In hypothesis testing, the power of a test is vital because it tells us the test's effectiveness in correctly rejecting a false null hypothesis. It represents the probability of avoiding a Type II error, meaning the likelihood that we correctly recognize a real difference when there is one. The greater the power, the better the test's ability to detect actual changes in the null hypothesis conditions. For the toll road study, this is the probability that we correctly determine the proportion of supporters is less than 30% when it truly is 20%. Power is calculated as 1 minus the Type II error probability, offering a measure of the test's reliability. Strong power means more confidence that the test correctly identifies the true proportion of supporters.

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Most popular questions from this chapter

The standard deviation of critical dimension thickness in semiconductor manufacturing is \(\sigma=20 \mathrm{nm}\). (a) State the null and alternative hypotheses used to demonstrate that the standard deviation is reduced. (b) Assume that the previous test does not reject the null hypothesis. Does this result provide strong evidence that the standard deviation has not been reduced? Explain.

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