/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 47 Let \(\mu\) denote the mean reac... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: 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
P-values: a) 0.0778, b) 0.1841, c) 0.0250, d) 0.0066, e) 0.5438.

Step by step solution

01

Understanding the Hypotheses

We are given the null hypothesis as \(H_0: \mu = 5\) and the alternative hypothesis as \(H_a: \mu > 5\). The alternative hypothesis indicates that we are conducting a right-tailed test since we are testing for mean values greater than 5.
02

Identifying the Test Statistic

The problem provides us with different \(z\) values for which we need to compute the \(P\)-value. These values are already calculated test statistics, so no further computation for these \(z\) values is needed in terms of standardizing.
03

P-Value Interpretation for Right-Tailed Test

For a right-tailed \(z\) test, the \(P\)-value is the probability that the standard normal random variable \(Z\) is greater than the observed \(z\) test statistic value. In statistical terms, this can be expressed as \(P(Z > z)\).
04

Finding P-Value for z = 1.42

Using standard normal distribution tables or statistical software, find the \(P\)-value for \(z = 1.42\): \[P(Z > 1.42) \approx 0.0778\]
05

Finding P-Value for z = 0.90

For \(z = 0.90\): \[P(Z > 0.90) \approx 0.1841\]
06

Finding P-Value for z = 1.96

For \(z = 1.96\): \[P(Z > 1.96) \approx 0.0250\]
07

Finding P-Value for z = 2.48

For \(z = 2.48\): \[P(Z > 2.48) \approx 0.0066\]
08

Finding P-Value for z = -0.11

For \(z = -0.11\): \[P(Z > -0.11) \approx 0.5438\] because the \(z\) value is less than the mean, indicating a large \(P\)-value.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Z-Test
The Z-Test is a statistical method used to determine if there's a significant difference between the sample mean and the population mean. It becomes highly useful when dealing with large sample sizes (typically over 30), as it assumes the sampling distribution of the sample mean is normally distributed.
  • One key aspect of the Z-Test is that it requires known population parameters, specifically the population standard deviation.
  • The test calculates a statistic known as the Z-score, which indicates how many standard deviations a sample mean is from the population mean.
The formula for the Z-score is: \[z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} \]Where \( \bar{x} \) is the sample mean, \( \mu \) is the population mean, \( \sigma \) is the population standard deviation, and \( n \) is the sample size.This test helps in making decisions about hypothesis testing, especially when testing detailed scenarios about population means.
P-Value
A P-Value is a crucial statistic in hypothesis testing that helps determine the significance of your results. It tells you the probability that the observed results would occur by random chance if the null hypothesis were true.
  • The P-Value is a probability, ranging from 0 to 1. Smaller P-Values indicate stronger evidence against the null hypothesis.
  • A commonly used significance level to evaluate P-Values is 0.05. If the P-Value is less than 0.05, the null hypothesis is often rejected in favor of the alternative hypothesis.
It鈥檚 important to interpret the P-Value correctly in the context of your test. For a right-tailed test, the P-Value is calculated as the probability of observing a test statistic as extreme as, or more extreme than, what was observed, if the null hypothesis is true.
Right-Tailed Test
A Right-Tailed Test is a type of hypothesis test that looks for evidence that a population parameter is greater than a hypothesized value. This implies we are focused on the right side of the probability distribution of the test statistic.
  • In these tests, we set up an alternative hypothesis that the parameter is greater than the null hypothesis value.
  • We are interested in the probability of the test statistic falling in the upper tail of the distribution.
For example, if our alternative hypothesis states that \( \mu > 5 \), we will calculate the probability of getting a test statistic that is greater than what we found: \( P(Z > z) \).
A right-tailed test is often employed when the one-sided nature of the hypothesis is clear, meaning we鈥檙e specifically checking for an increase in a parameter.
Statistical Hypotheses
Statistical Hypotheses are statements about the population parameters that we want to test. Generally, we deal with two types of hypotheses: the null hypothesis and the alternative hypothesis.
  • The null hypothesis (denoted as \( H_0 \)) is a statement asserting that there is no effect or difference. It often contains an equality, such as \( \mu = 5 \).
  • The alternative hypothesis (denoted as \( H_a \)) challenges the null hypothesis by suggesting a different outcome or effect, such as \( \mu > 5 \).
Hypothesis testing involves using sample data to assess the plausibility of these statements. We use tests such as Z-Tests to decide whether to reject or fail to reject the null hypothesis based on a calculated test statistic and corresponding P-Value. These hypotheses guide the direction and methodology of the testing process.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Give as much information as you can about the \(P\)-value of a \(t\) test in each of the following situations: a. Upper-tailed test, \(\mathrm{df}=8, t=2.0\) b. Lower-tailed test, \(\mathrm{df}=11, t=-2.4\) c. Two-tailed test, \(\mathrm{df}=15, t=-1.6\) d. Upper-tailed test, df \(=19, t=-.4\) e. Upper-tailed test, df \(=5, t=5.0\) f. Two-tailed test, df \(=40, t=-4.8\)

The calibration of a scale is to be checked by weighing a 10-kg test specimen 25 times. Suppose that the results of different weighings are independent of one another and that the weight on each trial is normally distributed with \(\sigma=.200 \mathrm{~kg}\). Let \(\mu\) denote the true average weight reading on the scale. a. What hypotheses should be tested? b. Suppose the scale is to be recalibrated if either \(\bar{x} \geq 10.1032\) or \(\bar{x} \leq 9.8968\). What is the probability that recalibration is carried out when it is actually unnecessary? c. What is the probability that recalibration is judged unnecessary when in fact \(\mu=10.1\) ? When \(\mu=9.8\) ? d. Let \(z=(\bar{x}-10) /(\sigma / \sqrt{n})\). For what value \(c\) is the rejection region of part (b) equivalent to the "two-tailed" region either \(z \geq\) cor \(z \leq-c\) ? e. If the sample size were only 10 rather than 25 , how should the procedure of part (d) be altered so that \(\alpha=.05\) ? f. Using the test of part (e), what would you conclude from the following sample data? \(\begin{array}{lllll}9.981 & 10.006 & 9.857 & 10.107 & 9.888 \\ 9.728 & 10.439 & 10.214 & 10.190 & 9.793\end{array}\) g. Re-express the test procedure of part (b) in terms of the standardized test statistic \(Z=(\bar{X}-10) /(\sigma / \sqrt{n)} .\)

The melting point of each of 16 samples of a brand of hydrogenated vegetable oil was determined, resulting in \(\bar{x}=94.32\). Assume that the distribution of melting point is normal with \(\sigma=1.20\). a. Test \(H_{0}: \mu=95\) versus \(H_{\mathrm{a}}: \mu \neq 95\) using a two- tailed level .01 test. b. If a level \(.01\) test is used, what is \(\beta(94)\), the probability of a type II error when \(\mu=94\) ? c. What value of \(n\) is necessary to ensure that \(\beta(94)=.1\) when \(\alpha=.01 ?\)

The article "Statistical Evidence of Discrimination" (J. Amer. Statist. Assoc., 1982: 773-783) discusses the court case Swain v. Alabama (1965), in which it was alleged that there was discrimination against blacks in grand jury selection. Census data suggested that \(25 \%\) of those eligible for grand jury service were black, yet a random sample of 1050 people called to appear for possible duty yielded only 177 blacks. Using a level \(.01\) test, does this data argue strongly for a conclusion of discrimination?

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.
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Geometry

Read Explanation

Logic and Functions

Read Explanation

Statistics

Read Explanation

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.