/*! 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 50 Newly purchased tires of a certa... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 50

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. \(-5.3\)

Short Answer

Expert verified
a. 0.0358, b. 0.0802, c. 0.5824, d. 0.1586, e. 0.

Step by step solution

01

Understanding the Problem

We need to compute the P-values for various given values of the test statistic \(z\), specifically \(z = 2.10, -1.75, -0.55, 1.41, -5.3\). The test is for a two-tailed hypothesis test, where the null hypothesis states that the population mean \(\mu = 30\), and the alternative hypothesis is \(\mu eq 30\).
02

Locate Critical Regions

For a two-tailed test, the P-value is computed as \(P = 2P(Z > |z|)\). This means we will first find the probability in the right tail for the absolute value of each \(z\) value and then double it, as the hypothesis is not directional.
03

Use Z-Table for P-value Calculation

Utilize a standard normal distribution (Z-table) to find the probability for each absolute \(z\) value. - For \(z = 2.10\), locate \(P(Z > 2.10)\), - For \(z = -1.75\), locate \(P(Z > 1.75)\), - For \(z = -0.55\), locate \(P(Z > 0.55)\), - For \(z = 1.41\), locate \(P(Z > 1.41)\), - For \(z = -5.3\), locate \(P(Z > 5.3)\).
04

Calculate individual P-values

Using a Z-table:- For \(z = 2.10\), \(P(Z > 2.10) = 0.0179\), hence \(P = 2 \times 0.0179 = 0.0358\).- For \(z = -1.75\), \(P(Z > 1.75) = 0.0401\), hence \(P = 2 \times 0.0401 = 0.0802\).- For \(z = -0.55\), \(P(Z > 0.55) = 0.2912\), hence \(P = 2 \times 0.2912 = 0.5824\).- For \(z = 1.41\), \(P(Z > 1.41) = 0.0793\), hence \(P = 2 \times 0.0793 = 0.1586\).- For \(z = -5.3\), \(P(Z > 5.3)\) is very small, nearly zero, hence \(P \approx 0.\)

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.

Hypothesis Testing
Hypothesis testing is a fundamental technique in statistics used to make inferences about populations based on sample data. The primary goal is to determine whether there is enough statistical evidence to support or reject a given hypothesis. In our exercise, we are dealing with two hypotheses:
  • The null hypothesis (\(H_0\)) posits that the true average pressure, denoted by \(\mu\), equals 30 (\(\mu = 30\)).
  • The alternative hypothesis (\(H_a\)) suggests that the true average pressure is not equal to 30 (\(\mu eq 30\)).
Hypothesis testing procedures often involve determining the validity of the null hypothesis. This is achieved by calculating a statistic (in this case, the \(z\)-statistic) from the sample data and comparing it with a known probability distribution to ascertain the probability of observing such a result if the null hypothesis were true.
Z-Statistic
The \(z\)-statistic is a measure that tells us how many standard deviations away a sample mean is from the population mean, assuming the null hypothesis is true. The calculation of the \(z\)-statistic aids in translating a sample observation (like tire pressure) into a standardized value, allowing us to utilize statistical tables easily.
  • A positive \(z\)-statistic indicates that the sample mean is above the population mean.
  • A negative \(z\)-statistic implies that the sample mean is below the population mean.
In the exercise given, we're provided with several \(z\)-values (e.g., 2.10, -1.75), and our task is to assess how extreme each of these values is to inform our hypothesis testing.
Normal Distribution
A normal distribution is a continuous probability distribution characterized by a symmetrical, bell-shaped curve. It is completely described by its mean (\(\mu\)) and standard deviation (\(\sigma\)).
  • The mean is located at the center of the distribution, and the standard deviation determines the spread.
  • Most data points fall within one standard deviation from the mean.
In hypothesis testing, particularly with \(z\)-tests, we assume that the sample statistic follows a normal distribution. Using a table for the standard normal distribution, we find probabilities associated with a \(z\)-score, which aids in determining the P-value during hypothesis testing.
Two-Tailed Test
A two-tailed test is used when we are interested in deviations in both directions from the hypothesized parameter value. This means we are testing whether the population parameter is significantly greater or less than the hypothesized value.
  • In the context of our exercise, we are evaluating if \(\mu\) is either less than or greater than 30, which means both ends of the distribution are considered.
  • The P-value, in this case, is calculated by doubling the probability obtained from the normal distribution table for the absolute value of the \(z\)-score.
This doubling of probability ensures that we capture the significance of the statistic on both the extreme ends of the distribution, facilitating a thorough hypothesis testing procedure.

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

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. \(-.11\)

Because of variability in the manufacturing process, the actual yielding point of a sample of mild steel subjected to increasing stress will usually differ from the theoretical yielding point. Let \(p\) denote the true proportion of samples that yield before their theoretical yielding point. If on the basis of a sample it can be concluded that more than \(20 \%\) of all specimens yield before the theoretical point, the production process will have to be modified. a. If 15 of 60 specimens yield before the theoretical point, what is the \(P\)-value when the appropriate test is used, and what would you advise the company to do? b. If the true percentage of "early yields" is actually \(50 \%\) (so that the theoretical point is the median of the yield distribution) and a level \(.01\) test is used, what is the probability that the company concludes a modification of the process is necessary?

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 \(n=10\) 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)?

For the following pairs of assertions, indicate which do not comply with our rules for setting up hypotheses and why (the subscripts 1 and 2 differentiate between quantities for two different populations or samples): a. \(H_{0}: \mu=100, H_{\mathrm{a}}: \mu>100\) b. \(H_{0}: \sigma=20, H_{\mathrm{a}}: \sigma \leq 20\) c. \(H_{0}: p \neq .25, H_{\mathrm{a}}: p=.25\) d. \(H_{0}: \mu_{1}-\mu_{2}=25, H_{\mathrm{a}}: \mu_{1}-\mu_{2}>100\) e. \(H_{0}: S_{1}^{2}=S_{2}^{2}, H_{\mathrm{a}}: S_{1}^{2} \neq S_{2}^{2}\) f. \(H_{0}: \mu=120, H_{\mathrm{a}}: \mu=150\) g. \(H_{0}: \sigma_{1} / \sigma_{2}=1, H_{\mathrm{a}}: \sigma_{1} / \sigma_{2} \neq 1\) h. \(H_{0}: p_{1}-p_{2}=-.1, H_{\mathrm{a}}: p_{1}-p_{2}<-.1\)

Have you ever been frustrated because you could not get a container of some sort to release the last bit of its contents? The article "Shake, Rattle, and Squeeze: How Much Is Left in That Container?" (Consumer Reports, May 2009: 8) reported on an investigation of this issue for various consumer products. Suppose five \(6.0 \mathrm{oz}\) tubes of toothpaste of a particular brand are randomly selected and squeezed until no more toothpaste will come out. Then each tube is cut open and the amount remaining is weighed, resulting in the following data (consistent with what the cited article reported): \(.53, .65, .46, .50, .37\). Does it appear that the true average amount left is less than \(10 \%\) of the advertised net contents? a. Check the validity of any assumptions necessary for testing the appropriate hypotheses. b. Carry out a test of the appropriate hypotheses using a significance level of \(.05\). Would your conclusion change if a significance level of .01 had been used? c. Describe in context type I and II errors, and say which error might have been made in reaching a conclusion.
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Discrete Mathematics

Read Explanation

Geometry

Read Explanation

Applied Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

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.