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WP 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. 0.9669; c. 0.3446

Step by step solution

01

Understand the Hypothesis Test

We have a one-sided hypothesis test where the null hypothesis is \( H_0: \mu = 10 \) and the alternative hypothesis is \( H_1: \mu > 10 \). The test statistic \( z_0 \) is given, and our goal is to calculate the \( P \)-value for each scenario to determine how extreme our sample results are under the null hypothesis.
02

Step 2a: Calculate the P-value for \( z_0 = 2.05 \)

The \( P \)-value is the probability that the test statistic is at least as extreme as that observed, under the null hypothesis. For \( z_0 = 2.05 \), this means finding \( P(Z > 2.05) \), where \( Z \) is a standard normal variable.Using a standard normal distribution table or calculator, we find:\[ P(Z > 2.05) = 1 - P(Z < 2.05) = 1 - 0.9798 = 0.0202 \]
03

Step 2b: Calculate the P-value for \( z_0 = -1.84 \)

For \( z_0 = -1.84 \), we look at \( P(Z > -1.84) \). Since this is a one-sided test and the alternative hypothesis looks for values greater than the null, calculate opposite cumulative probability:\[ P(Z > -1.84) = 1 - P(Z < -1.84) = 1 - 0.0331 = 0.9669 \]
04

Step 2c: Calculate the P-value for \( z_0 = 0.4 \)

For \( z_0 = 0.4 \), calculate \( P(Z > 0.4) \):Using the standard normal table or calculator:\[ P(Z > 0.4) = 1 - P(Z < 0.4) = 1 - 0.6554 = 0.3446 \]

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Key Concepts

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

P-value Calculation
P-value calculation is a key concept in hypothesis testing. The P-value tells us the probability of obtaining a test statistic as extreme as the one actually observed, assuming that the null hypothesis is true. In simpler terms, it helps determine if there's enough evidence to reject the null hypothesis.

To calculate the P-value in the context of a standard normal distribution, we often rely on:
  • Standard Normal Distribution Tables
  • Calculators or software tools
When given a test statistic like the values in points a, b, and c of the original exercise, you first determine how extreme these are. This involves using the standard normal distribution to find the probability that a standard normal variable exceeds the given test statistic. This probability is the P-value in a one-sided test where the alternative hypothesis specifies a direction (greater than, in this case).

For example, for a test statistic of 2.05, the P-value is calculated by finding the probability that a standard normal variable is greater than 2.05, leading to a P-value of 0.0202. This small P-value suggests that such an extreme result is rare under the null hypothesis.
Standard Normal Distribution
The standard normal distribution is a critical part of calculating P-values and understanding hypothesis tests. It's a normal distribution with a mean of 0 and a standard deviation of 1. This is often denoted as a Z-distribution.

Key properties of the standard normal distribution include:
  • Symmetrical about the mean
  • Mean = 0, Standard Deviation = 1
  • Total area under the curve equals 1
The Z-distribution allows us to determine the probability of obtaining a value in any particular range, utilizing Z-scores, which measure how many standard deviations away from the mean a value falls. With Z-tables or calculators, these probabilities are easy to discover.

In our exercise, each given test statistic (e.g., 2.05, -1.84, 0.4) is actually a Z-score. With it, we compute what area lies to its right (in one-sided tests) to find the P-value, which reflects how extreme the data is under the normal distribution.
One-Sided Tests
One-sided tests, as encountered in our original exercise, are hypothesis tests where the alternative hypothesis specifies a particular direction of interest. In the provided example, the hypothesis test checks if the mean (\( \mu \)) is greater than 10.

Key features of one-sided tests include:
  • Directional alternative hypothesis (e.g., \( H_1: \mu > 10 \) )
  • Focuses on deviations in a single direction
  • Calculates the probability of observing a test statistic more extreme in that specified direction
In practical terms, one-sided tests are useful when we have a clear hypothesis that a parameter (like the mean) will only go up or down. With these tests, the P-value is the probability that a test statistic is greater than or less than the observed one, only considering the extreme values in the specified direction.

In summary, for a test statistic of 2.05 from our exercise, the essence of the one-sided test is computing how likely it is to get such a value or one even bigger, assuming the null hypothesis is true. This approach helps in identifying significant deviations that support the alternative hypothesis in a particular direction.

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Most popular questions from this chapter

For the hypothesis test \(H_{0}: \mu=7\) against \(H_{1}: \mu \neq 7\) with variance unknown and \(n=20,\) approximate the \(P\) -value for each of the following test statistics. a. \(t_{0}=2.05\) b. \(t_{0}=-1.84\) c. \(t_{0}=0.4\)

A consumer products company is formulating a new shampoo and is interested in foam height (in millimeters). Foam height is approximately normally distributed and has a standard deviation of 20 millimeters. The company wishes to test \(H_{0}: \mu=175\) millimeters versus \(H_{1}: \mu>175\) millimeters, using the results of \(n=10\) samples. a. Find the type I error probability \(\alpha\) if the critical region is \(\bar{x}>185\) b. What is the probability of type II error if the true mean foam height is 185 millimeters? c. Find \(\beta\) for the true mean of 195 millimeters.

Medical researchers have developed a new artificial heart constructed primarily of titanium and plastic. The heart will last and operate almost indefinitely once it is implanted in the patient's body, but the battery pack needs to be recharged about every 4 hours. A random sample of 50 battery packs is selected and subjected to a life test. The average life of these batteries is 4.05 hours. Assume that battery life is normally distributed with standard deviation \(\sigma=0.2\) hour. a. Is there evidence to support the claim that mean battery life exceeds 4 hours? Use \(\alpha=0.05 .\) b. What is the \(P\) -value for the test in part (a)? c. Compute the power of the test if the true mean battery life is 4.5 hours. d. What sample size would be required to detect a true mean battery life of 4.5 hours 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 one-sided confidence bound on the mean life.

The mean bond strength of a cement product must be at least 10000 psi. The process by which this material is manufactured must show equivalence to this standard. If the process can manufacture cement for which the mean bond strength is at least 9750 psi, it will be considered equivalent to the standard. A random sample of six observations is available, and the sample mean and standard deviation of bond strength are 9360 psi and 42.6 psi, respectively. a. State the appropriate hypotheses that must be tested to demonstrate equivalence. b. What are your conclusions using \(\alpha=0.05 ?\)

Suppose that 10 sets of hypotheses of the form $$H_{0}: \mu=\mu_{0} \quad H_{1}: \mu \neq \mu_{0}$$ have been tested and that the \(P\) -values for these tests are 0.12 , \(0.08 .0 .93,0.02,0.01,0.05,0.88,0.15,0.13,\) and \(0.06 .\) Use Fisher's procedure to combine all of these \(P\) -values. What conclusions can you draw about these hypotheses?

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