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For the hypothesis test \(H_{0}: \mu=10\) against \(H_{1}: \mu>10\) and variance known, calculate the \(P\) -value for each of the following test statistics. (a) \(z_{0}=2.05\) (b) \(z_{0}=-1.84\) (c) \(z_{0}=0.4\)

Short Answer

Expert verified
(a) 0.0202, (b) 0.9671, (c) 0.3446

Step by step solution

01

Understand the Hypothesis Test

The hypothesis test is for the population mean. We have a null hypothesis \(H_0: \mu = 10\) and an alternative hypothesis \(H_1: \mu > 10\). This is a one-tailed test to determine if the population mean is greater than 10.
02

Significance from Test Statistics

For a statistical test with known variance, the standard normal distribution (z-distribution) is used. The test statistic \(z_0\) tells us how many standard deviations our statistic is away from the mean. We need to find the p-value associated with each \(z_0\).
03

Calculate P-value for \(z_{0}=2.05\)

To find the p-value, use the standard normal distribution. Look up \(z=2.05\) in the z-table or use a calculator to find the probability that \(Z > 2.05\). The p-value is the area to the right of 2.05. \[ P(Z > 2.05) \approx 0.0202 \]
04

Calculate P-value for \(z_{0}=-1.84\)

Even though \(z_0 = -1.84\) indicates a value less than the mean under the null, in the context of this \(H_1: \mu > 10\) test, we consider probabilities to the right of this cutoff, which gives a p-value of more than 0.5. \[ P(Z > -1.84) \approx 0.9671 \]
05

Calculate P-value for \(z_{0}=0.4\)

Look up or compute \(P(Z > 0.4)\) in a similar way. Since 0.4 is right-side of the mean, this p-value measures the probability \(Z > 0.4\). \[ P(Z > 0.4) \approx 0.3446 \]
06

Interpret the Results

P-values indicate the probability of observing a test statistic as extreme as \(z_0\), given \(H_0\) is true. Lower p-values suggest stronger evidence against \(H_0\). Thus, \(z_0 = 2.05\) gives the strongest evidence against \(H_0\), as its p-value is lowest.

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Key Concepts

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

Z-distribution
The Z-distribution, or standard normal distribution, is at the heart of many hypothesis tests. This distribution is symmetric and has a bell-shaped curve. The mean of the Z-distribution is 0, and the standard deviation is 1. Therefore, any z-score you calculate describes the number of standard deviations a data point is from the mean.
In hypothesis testing, this is instrumental for comparing how far a sample statistic deviates from a population parameter, under the null hypothesis.
  • **Mean**: The Z-distribution has a mean of 0.
  • **Standard Deviation**: The standard deviation is 1, making it "standardized".
  • **Symmetry**: It is symmetrical around the mean.
When using a Z-distribution for hypothesis testing, it helps determine where a test statistic falls compared to the expected normal distribution. The further away from 0 (either positively or negatively), the more unusual your result is under the null hypothesis, suggesting potential significance.
P-value calculation
P-value calculation is critical in hypothesis testing as it quantifies the evidence against the null hypothesis. To calculate a p-value using the z-distribution, follow these steps:
1. **Locate Your Test Statistic**: First, determine the z-score corresponding to your data. The z-score indicates how many standard deviations away from the population mean your result is.
2. **Use the Z-table or Calculator**: With the z-score in hand, you can find the probability associated with this score by using a Z-table or calculator. This gives you the area under the curve to the left of your z-score.
3. **Calculate the P-value**: In a one-tailed test, like in the problem, the p-value is the area to the one end of the z-score. If looking for the area to the right, as in this exercise, you use: - **P-value for positive z-score**: Subtract the Z-table value from 1. - **P-value for negative z-score**: Use the Z-table value directly, or consider the right-tail probability which is often simplifying with 1 - p. A small p-value indicates strong evidence against the null hypothesis, signaling that the observation is rare under the assumption that the null hypothesis is true.
One-tailed test
A one-tailed test is used in hypothesis testing when the alternative hypothesis specifies a direction of the effect. In the given problem, we're checking if the mean is significantly greater than a specific value (10).
**Advantages of One-tailed Tests**: - **Directional Hypothesis**: Helps when you have a clear expectation that effects will only go one way. - **More Power**: For the same significance level, it offers more power to detect an effect in the specified direction compared to a two-tailed test.
The significance of a one-tailed test is determined by considering probabilities in one direction from the mean:
  • **Right-tailed test**: If we're testing for 'greater than' scenarios, the p-value from the Z-distribution reflects the probability of observing a test statistic as extreme as calculated or more, in the right tail of the distribution.
  • **Left-tailed test**: The reverse is true if hypothetically we were testing for 'less than' scenarios, examining the left tail.
When using a one-tailed test, select the direction wisely as it assumes there can't be a significant result if the effect goes the opposite way. Therefore, one-tailed tests should only be used when you have a strong theory or prior evidence suggesting the direction of the effect.

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Most popular questions from this chapter

State whether each of the following situations is a correctly stated hypothesis testing problem and why. (a) \(H_{0}: \mu=25, H_{1}: \mu \neq 25\) (b) \(H_{0}: \sigma>10, H_{1}: \sigma=10\) (c) \(H_{0}: \bar{x}=50, H_{1}: \bar{x} \neq 50\) (d) \(H_{0}: p=0.1, H_{1}: p=0.5\) (e) \(H_{0}: s=30, H_{1}: s>30\)

Consider the following frequency table of observations on the random variable \(X\) : $$\begin{array}{llrrrr}\text { Values } & 0 & 1 & 2 & 3 & 4 \\\\\text { Frequency } & 4 & 21 & 10 & 13 & 2\end{array}$$ (a) Based on these 50 observations, is a binomial distribution with \(n=6\) and \(p=0.25\) an appropriate model? Perform a goodness-of-fit procedure with \(\alpha=0.05 .\) (b) Calculate the \(P\) -value for this test.

Consider the test of \(H_{0}: \sigma^{2}=5\) against \(H_{1}: \sigma^{2}<5 .\) Approximate the \(P\) -value for each of the following test statistics. (a) \(x_{0}^{2}=25.2\) and \(n=20\) (b) \(x_{0}^{2}=15.2\) and \(n=12\) (c) \(x_{0}^{2}=4.2\) and \(n=15\)

An article in Food Chemistry ["A Study of Factors Affecting Extraction of Peanut (Arachis Hypgaea L.) Solids with Water" (1991, Vol. \(42(2),\) pp. \(153-165)\) ] reported that the percent protein extracted from peanut milk as follows: $$\begin{array}{llllllll}78.3 & 77.1 & 71.3 & 84.5 & 87.8 & 75.7 & 64.8 & 72.5 \\\78.2 & 91.2 & 86.2 & 80.9 & 82.1 & 89.3 & 89.4 & 81.6\end{array}$$ (a) Can you support a claim that the mean percent protein extracted exceeds 80 percent? Use \(\alpha=0.05\) (b) Is there evidence that the percent protein extracted is normally distributed? (c) What is the \(P\) -value of the test statistic computed in part (a)?

An article in the ASCE Journal of Energy Engineering (1999, Vol. \(125,\) pp. \(59-75\) ) describes a study of the thermal inertia properties of autoclaved aerated concrete used as a building material. Five samples of the material were tested in a structure, and the average interior temperatures \(\left({ }^{\circ} \mathrm{C}\right)\) reported were as follows: \(23.01,22.22,22.04,22.62,\) and \(22.59 .\) (a) Test the hypotheses \(H_{0}: \mu=22.5\) versus \(H_{1}: \mu \neq 22.5,\) using \(\alpha=0.05 .\) Find the \(P\) -value. (b) Check the assumption that interior temperature is normally distributed. (c) Compute the power of the test if the true mean interior temperature is as high as 22.75 . (d) What sample size would be required to detect a true mean interior temperature as high as 22.75 if you wanted the power of the test to be at least \(0.9 ?\) (e) Explain how the question in part (a) could be answered by constructing a two-sided confidence interval on the mean interior temperature.

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