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Chapter 8: Problem 47

Let \(\mu\) denote the mean reaction time to a certain stimulus. For a large- sample \(z\) test of \(H_{0}: \mu=5\) versus \(H_{\mathrm{a}}: \mu>5\), find the \(P\)-value associated with each of the given values of the \(z\) test statistic. a. \(1.42\) b. 90 c. \(1.96\) d. \(2.48\) e. \(-.11\)

Short Answer

Expert verified
a. 0.0778, b. 0, c. 0.0250, d. 0.0066, e. 0.5438

Step by step solution

01

Understanding the Problem

We are conducting a hypothesis test where the null hypothesis is \( H_0: \mu = 5 \) and the alternative hypothesis is \( H_a: \mu > 5 \). We are given different values of the \( z \) test statistic and need to find the corresponding \( P \)-value for each one in the context of a right-tailed test.
02

Identify the Test Type

This is a one-tailed test in the positive (right) direction, given by the alternative hypothesis \( H_a: \mu > 5 \). We're looking for the probability of observing a \( z \) value as extreme or more extreme than the given \( z \) value.
03

Calculate the P-Value for z = 1.42

For a \( z \) test statistic of 1.42, use a standard normal distribution table or calculator to find the area to the right of 1.42. This is the \( P \)-value for \( z = 1.42 \). The area to the right is approximately 0.0778.
04

Calculate the P-Value for z = 90

A \( z \) value of 90 is extraordinarily high, beyond typical standard normal table values. Thus, the \( P \)-value is essentially 0 or very close to 0, indicating an almost certain rejection of the null hypothesis.
05

Calculate the P-Value for z = 1.96

For a \( z \) test statistic of 1.96, the area to the right (\( P \)-value) is approximately 0.0250 using a standard normal table or calculator.
06

Calculate the P-Value for z = 2.48

For a \( z \) test statistic of 2.48, use a standard normal distribution table or calculator. The \( P \)-value is approximately 0.0066, indicating strong evidence against the null hypothesis.
07

Calculate the P-Value for z = -0.11

For a \( z \) test statistic of -0.11, the \( P \)-value is close to 0.5 (or 1 minus the area on the left for a two-tailed test), since we are looking in a positive direction and -0.11 provides no evidence of \( \mu > 5 \). This is around 0.5438.

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Key Concepts

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

P-value
The p-value is a core concept in hypothesis testing. It's the probability of obtaining a sample statistic as extreme as the test statistic, under the assumption that the null hypothesis is true. In simpler terms, the p-value helps us understand how strong our evidence is against the null hypothesis.
  • A small p-value (typically 鈮� 0.05) indicates strong evidence against the null hypothesis, so we reject it.
  • A large p-value (> 0.05) suggests weak evidence against the null hypothesis, so we fail to reject it.
  • A p-value close to 0.5 means the observed data is likely if the null hypothesis is true, showing no significant effect.
In the provided exercise, for each z value, we calculate the p-value to determine the strength of evidence against the null hypothesis. For instance, when the z-value is 1.42, the p-value is approximately 0.0778, suggesting marginal evidence against the null hypothesis.
z-test
The z-test is a statistical test used to determine if there is a significant difference between sample and population means when the variance is known and the sample size is large. It is based on the standard normal distribution. The test statistic in a z-test follows a normal distribution with a mean of 0 and a standard deviation of 1.
  • It compares the difference between the sample mean and the population mean, standardized by the standard error.
  • This test is appropriate when the sample size is large (usually n > 30).
  • It's particularly useful when the population variance is known.
In the exercise, the z-test helps us determine how far away each sample mean is from the population mean, under the hypothesis that their difference is due to random sampling variability.
Standard Normal Distribution
The standard normal distribution is a special case of the normal distribution with a mean of 0 and a standard deviation of 1. It is denoted as the Z-distribution. This distribution is symmetric around the mean, bell-shaped, and plays a crucial role in hypothesis testing.
  • The z-scores represent the number of standard deviations away from the mean of a data point.
  • Z-scores become especially useful because they allow different normal distributions to be converted to a standardized form, which aids in calculating probabilities.
  • In hypothesis tests, we use a standard normal distribution table to find areas under the curve, translating our z-score into a probability (p-value).
In our exercise, we look up z-scores in the standard normal distribution to find corresponding p-values for each given z-score.
Alternative Hypothesis
The alternative hypothesis (\( H_a \)) is a statement that contradicts the null hypothesis (\( H_0 \)). It is what you aim to provide evidence for through your testing. It can take different forms, depending on the research question, and is typically expressed using inequalities.
  • In a one-tailed test, the alternative hypothesis posits whether a parameter is greater than or less than the null hypothesis value.
  • In a two-tailed test, the alternative hypothesis suggests that a parameter differs in either direction from the null hypothesis value.
  • The alternative hypothesis is proven when the p-value is low enough to reject the null hypothesis.
In our problem, the alternative hypothesis is that the mean reaction time is greater than 5 (\( H_a: \, \mu > 5 \)). This directional hypothesis leads us to perform a right-tailed test.

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Most popular questions from this chapter

Newly purchased tires of a certain type are supposed to be filled to a pressure of \(30 \mathrm{lb} / \mathrm{in}^{2}\). Let \(\mu\) denote the true average pressure. Find the \(P\)-value associated with each given \(z\) statistic value for testing \(H_{0}: \mu=30\) versus \(H_{\mathrm{a}}: \mu \neq 30\). a. \(2.10\) b. \(-1.75\) c. \(-.55\) d. \(1.41\) e. \(-5.3\)

A mixture of pulverized fuel ash and Portland cement to be used for grouting should have a compressive strength of more than \(1300 \mathrm{KN} / \mathrm{m}^{2}\). The mixture will not be used unless experimental evidence indicates conclusively that the strength specification has been met. Suppose compressive strength for specimens of this mixture is normally distributed with \(\sigma=60\). Let \(\mu\) denote the true average compressive strength. a. What are the appropriate null and alternative hypotheses? b. Let \(\bar{X}\) denote the sample average compressive strength for \(n=20\) randomly selected specimens. Consider the test procedure with test statistic \(\bar{X}\) and rejection region \(\bar{x} \geq 1331.26\). What is the probability distribution of the test statistic when \(H_{0}\) is true? What is the probability of a type I error for the test procedure? c. What is the probability distribution of the test statistic when \(\mu=1350\) ? Using the test procedure of part (b), what is the probability that the mixture will be judged unsatisfactory when in fact \(\mu=1350\) (a type II error)? d. How would you change the test procedure of part (b) to obtain a test with significance level .05? What impact would this change have on the error probability of part (c)? e. Consider the standardized test statistic \(Z=\) \((\bar{X}-1300) /(\sigma / \sqrt{n})=(\bar{X}-1300) / 13.42\). What are the values of \(Z\) corresponding to the rejection region of part (b)?

Let the test statistic \(Z\) have a standard normal distribution when \(H_{0}\) is true. Give the significance level for each of the following situations: a. \(H_{\mathrm{a}}: \mu>\mu_{0}\), rejection region \(z \geq 1.88\) b. \(H_{\mathrm{a}}: \mu<\mu_{0}\), rejection region \(z \leq-2.75\) c. \(H_{\mathrm{a}}: \mu \neq \mu_{0}\), rejection region \(z \geq 2.88\) or \(z \leq-2.88\)

State DMV records indicate that of all vehicles undergoing emissions testing during the previous year, \(70 \%\) passed on the first try. A random sample of 200 cars tested in a particular county during the current year yields 124 that passed on the initial test. Does this suggest that the true proportion for this county during the current year differs from the previous statewide proportion? Test the relevant hypotheses using \(\alpha=.05\).

Water samples are taken from water used for cooling as it is being discharged from a power plant into a river. It has been determined that as long as the mean temperature of the discharged water is at most \(150^{\circ} \mathrm{F}\), there will be no negative effects on the river's ecosystem. To investigate whether the plant is in compliance with regulations that prohibit a mean discharge-water temperature above \(150^{\circ}, 50\) water samples will be taken at randomly selected times, and the temperature of each sample recorded. The resulting data will be used to test the hypotheses \(H_{0}: \mu=150^{\circ}\) versus \(H_{\mathrm{a}}: \mu>150^{\circ}\). In the context of this situation, describe type I and type II errors. Which type of error would you consider more serious? Explain.
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