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You have an SRS of size \(n=25\) from a Normal distribution with \(\sigma=2.5\). You wish to test $$ \begin{aligned} &H_{0}: \mu=0 \\ &H_{a}: \mu>0 \end{aligned} $$ You decide to reject \(H_{0}\) if \(x^{-} \bar{x}>0\) and not reject \(H_{0}\) otherwise. (a) Find the probability of a Type I error. That is, find the probability that the test rejects \(H_{0}\) when in fact \(\mu=0\). (b) Find the probability of a Type II error when \(\mu=0.5\). This is the probability that the test fails to reject \(H_{0}\) when in fact \(\mu=0.5\). (c) Find the probability of a Type II error when \(\mu=1.0\).

Short Answer

Expert verified
(a) 0.5; (b) 0.1587; (c) 0.0228.

Step by step solution

01

Identify Type I Error

A Type I error is rejecting the null hypothesis \(H_0\) when it is true. For our hypothesis test, we will reject \(H_0\) if \(\bar{x} > 0\). Since \(\mu = 0\) under the null hypothesis, and \(\bar{x}\) follows a Normal distribution \(N(0, \frac{\sigma}{\sqrt{n}})\) when \(\mu=0\), where \(\sigma=2.5\) and \(n=25\). This means \(\bar{x} \sim N(0, 0.5)\). Therefore, the probability of a Type I error is \(P(\bar{x} > 0)\).
02

Calculate Probability of Type I Error

The probability \(P(\bar{x} > 0)\) when \(\bar{x} \sim N(0, 0.5)\) is determined by standardizing. We calculate the z-score: \(z = \frac{0 - 0}{0.5} = 0\). Look up this value in the standard Normal distribution table. The probability that \(Z > 0\) is 0.5. Therefore, the probability of a Type I error is 0.5.
03

Identify Type II Error When \(\mu = 0.5\)

A Type II error is failing to reject \(H_0\) when it is false. We fail to reject \(H_0\) if \(\bar{x} \leq 0\). Now, assume \(\mu=0.5\). In this case, \(\bar{x} \sim N(0.5, 0.5)\). We calculate \(P(\bar{x} \leq 0)\), the probability of a Type II error when \(\mu=0.5\).
04

Calculate Probability of Type II Error When \(\mu = 0.5\)

For \(\bar{x} \sim N(0.5, 0.5)\), calculate the z-score for \(\bar{x} = 0\): \(z = \frac{0 - 0.5}{0.5} = -1\). Use the standard Normal distribution table to find \(P(Z \leq -1)\), which is approximately 0.1587. Therefore, the probability of a Type II error when \(\mu=0.5\) is 0.1587.
05

Identify Type II Error When \(\mu = 1.0\)

Similarly, for \(\mu=1.0\), \(\bar{x} \sim N(1.0, 0.5)\). The Type II error probability is \(P(\bar{x} \leq 0)\), where we assume \(\mu=1.0\). Calculate the corresponding z-score and probability.
06

Calculate Probability of Type II Error When \(\mu = 1.0\)

Calculate the z-score for \(\bar{x} = 0\): \(z = \frac{0 - 1.0}{0.5} = -2\). From the standard Normal distribution table, \(P(Z \leq -2)\) is approximately 0.0228. Thus, the probability of a Type II error when \(\mu=1.0\) is 0.0228.

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Key Concepts

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

Type I Error
A Type I error occurs when we reject the null hypothesis (H_0) even though it is true. It's like crying wolf when there is no wolf! Imagine setting a rule that we reject H_0 if \( \bar{x} > 0 \). Here, the null hypothesis is \( \mu = 0 \), and if the sample mean \( \bar{x} \) follows a Normal distribution with a mean 0 and standard deviation 0.5, as given by \( N(0, 0.5) \), we're dealing with a Type I error every time we mistakenly reject H_0 based on our rule.
To find the probability of a Type I error, we calculate \( P(\bar{x} > 0) \). For a standardized random variable following a standard Normal distribution, this is just looking at where the probability starts. By calculating the z-score \( z = (0 - 0) / 0.5 = 0 \) and checking a standard Normal distribution table, we find that \( P(Z > 0) = 0.5 \). Essentially, there's a 50% chance of mistakenly rejecting H_0. This can be a high risk if improperly accounted for in critical real-world applications, making understanding and controlling Type I errors crucial in hypothesis tests.
Type II Error
Type II error concerns the opposite scenario: when we fail to reject the null hypothesis (H_0) when it is actually false. It's like ignoring the wolf when it really is lurking nearby. In hypothesis testing, ensuring this error type is minimized is essential to avoid missing true effects or differences.
For example, suppose \( \mu = 0.5 \), meaning the actual situation contradicts the null hypothesis. Here, \( \bar{x} \) would be expected to follow \( N(0.5, 0.5) \). To calculate the probability of a Type II error, denoted as \( \beta \), we find \( P(\bar{x} \leq 0) \). This requires calculating the z-score for the boundary point where we fail to reject H_0.
Use \( z = (0 - 0.5) / 0.5 = -1 \) and check the standard Normal distribution table to find \( P(Z \leq -1) \approx 0.1587 \). Hence, the likelihood of not detecting the true mean is about 15.87%. When \( \mu = 1.0 \), this chance decreases to 2.28% since the z-score changes to \( z = -2 \), producing \( P(Z \leq -2) = 0.0228 \). Recognizing and adjusting for Type II errors boosts the robustness of our conclusions.
Normal Distribution
The Normal distribution is a continuous probability distribution commonly used in statistics due to its appealing properties, like symmetry and the central peak. It's characterized by the mean (\mu), which determines the location of the center, and the standard deviation (\sigma), which measures spread or variability.
Its bell-shaped curve signifies that data near (\mu) is more frequent in occurrence than data far from (\mu). Mathematically, the probability density function of a Normal distribution is given by:
\[ f(x | \mu, \sigma) = \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{1}{2} \left(\frac{x-\mu}{\sigma}\right)^2} \]
When handling samples, the distribution of sample means (\bar{x}) from populations with a Normal distribution also forms a Normal distribution thanks to the central limit theorem, especially when the sample size is reasonably large, like 25 in our scenario. Understanding this concept assists in path and probability prediction and hypothesis testing.
Standard Normal Distribution
The standard Normal distribution is a special case of the Normal distribution where the mean is 0 and the standard deviation is 1. Also known as the z-distribution, it simplifies calculations and comparisons using z-scores, a method of standardizing data points into this universal scale.
A transformation using the z-score formula:
\[ z = \frac{x - \mu}{\sigma} \]
converts data following any Normal distribution into the standard Normal distribution. This standardization allows us to utilize well-established statistical tables to find probabilities and critical values quickly. These standard tables, or charts, provide cumulative probabilities associated with z-scores, making it easier to interpret results and draw conclusions.
This concept is fundamental for calculating Type I and Type II errors, as previously described, with z-scores enabling us to look up precise probability estimates and facilitate comparisons across different contexts.

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Most popular questions from this chapter

Suppose that scores on the mathematics part of the National Assessment of Educational Progress (NAEP) test for eighthgrade students follow a Normal distribution with standard deviation \(\sigma=110\). You want to estimate the mean score within \(\pm 10\) with \(90 \%\) confidence. How large an SRS of scores must you choose?

Statisticians prefer large samples. Describe briefly the effect of increasing the size of a sample (or the number of subjects in an experiment) on each of the following: (a) The \(P\)-value of a test, when \(H_{0}\) is false and all facts about the population remain unchanged as \(n\) increases. (b) (Optional) The power of a fixed level \(\alpha\) test, when \(\alpha\), the alternative hypothesis, and all facts about the population remain unchanged.

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