/*! 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 33 Make the following tests of hypo... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 33

Make the following tests of hypotheses. a. \(H_{0}: \mu=80, \quad H_{1}: \mu \neq 80, \quad n=33, \quad \bar{x}=76.5, \quad \sigma=15, \quad \alpha=.10\) b. \(H_{0}: \mu=32, \quad H_{1}: \mu<32, \quad n=75, \quad \bar{x}=26.5, \quad \sigma=7.4, \quad \alpha=.01\) c. \(H_{0}: \mu=55, \quad H_{1}: \mu>55, \quad n=40, \quad \bar{x}=60.5, \quad \sigma=4, \quad \alpha=.05\)

Short Answer

Expert verified
a. Fail to reject \(H_{0}: \mu=80\) with \(\alpha=.10\)\nb. Reject \(H_{0}: \mu=32\) with \(\alpha=.01\)\nc. Reject \(H_{0}: \mu=55\) with \(\alpha=.05\).

Step by step solution

01

Z-test Statistic Calculation

First, you calculate the z-test statistic for each case. The formula to calculate the z-test statistic is \[Z= \frac{\bar{x}-\mu}{\sigma/\sqrt{n}}\]. \n\na. \(Z= \frac{76.5-80}{15/\sqrt{33}} = -1.32\)\nb. \(Z= \frac{26.5-32}{7.4/\sqrt{75}} = -5.41\)\nc. \(Z= \frac{60.5-55}{4/\sqrt{40}} = 6.89\)
02

Z-critical Value Calculation

Second, determine the z-critical values using the given \(\alpha\).\n\na. For two-sided test, \(\alpha/2= 0.05\). Z-critical values are \(\pm 1.645\) (refer to Z-table).\nb. For left-tailed test, \(\alpha = 0.01\). Z-critical value is \(-2.33\). \nc. For right-tailed test, \(\alpha = 0.05\). Z-critical value is \(1.645\).
03

Decision Rule

Third, compare Z-test statistic with Z-critical value and make the decision for each case.\n\na. The value of Z (-1.32) lies between the critical region. Therefore, fail to reject \(H_{0}\).\nb. The value of Z (-5.41) lies in the critical region. Therefore, reject \(H_{0}\).\nc. The value of Z (6.89) lies in the critical region. Therefore, reject \(H_{0}\).

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.

z-test
The z-test is a statistical method used to determine if there is a significant difference between sample data and a population mean. We use it when the population variance is known and the sample size is large (usually more than 30). The formula for the z-test statistic is:
  • \(Z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}}\)
Here:
  • \(\bar{x}\) is the sample mean
  • \(\mu\) is the population mean
  • \(\sigma\) is the population standard deviation
  • \(n\) is the sample size
By calculating this statistic, we can determine how far the sample mean is from the population mean in standard deviation units.
z-critical value
The z-critical value is a point on the z-distribution that defines the boundary for the region where we reject the null hypothesis. It depends on the chosen significance level \(\alpha\), which is the probability of rejecting the null hypothesis when it is true. This value can be found in z-tables:
  • For a significance level \(\alpha\), the z-critical value divides the area of the distribution into acceptance and rejection regions.
  • In a two-tailed test, the critical values split the distribution into two tails.
Understanding the z-critical value helps us decide whether the observed data is statistically significant.
decision rule
The decision rule in hypothesis testing provides the guideline whether to reject or fail to reject the null hypothesis. It compares the z-test statistic to the z-critical value:
  • If the z-test statistic falls into the critical region (beyond the critical value), we reject \(H_0\).
  • If the z-test statistic does not fall into the critical region, we fail to reject \(H_0\).
This rule helps us determine if the test findings are significant enough to support the alternate hypothesis.
two-tailed test
A two-tailed test is used when the alternative hypothesis suggests that the parameter is different from the null hypothesis but does not indicate a direction. It equally tests both possibilities of the parameter being either higher or lower. For this test:
  • The significance level is split between both tails of the distribution.
  • The critical region is divided into two parts, making it essential to find two z-critical values.
  • Example: If \(\mu eq 80\), we are concerned about deviations in either direction from 80.
Two-tailed tests offer a more reflective assessment when directionality isn't specified.
left-tailed test
A left-tailed test is employed when the alternative hypothesis suggests the parameter is less than the null hypothesis value. Here, all the critical region lies on the left side of the distribution:
  • The focus is on values that are significantly lower.
  • The entire significance level \(\alpha\) is in the left tail.
  • Example: If \(\mu < 32\), we only consider the possibility that the mean is less than 32.
This test is useful for situations where a decrease or lower value is of particular interest.
right-tailed test
A right-tailed test is conducted when the alternative hypothesis indicates the parameter is greater than the null hypothesis. It's focused on assessing the higher side of the distribution:
  • The critical region is in the right tail.
  • The entire significance level \(\alpha\) is on this side.
  • Example: If \(\mu > 55\), we are interested in whether the mean is significantly more than 55.
This test is suitable where an increase or higher value is the main concern.

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

A May 8,2008 , report on National Public Radio (www.npr.org) noted that the average age of firsttime mothers in the United States is slightly higher than 25 years. Suppose that a recently taken random sample of 57 first-time mothers from Missouri produced an average age of \(23.90\) years and that the population standard deviation is known to be \(4.80\) years. a. Find the \(p\) -value for the test of hypothesis with the alternative hypothesis that the current mean age of all first-time mothers in Missouri is less than 25 years. Will you reject the null hypothesis at \(\alpha=.025\) ? b. Test the hypothesis of part a using the critical-value approach and \(\alpha=.025\).

Brooklyn Corporation manufactures DVDs. The machine that is used to make these DVDs is known. to produce not more than \(5 \%\) defective DVDs. The quality control inspector selects a sample of 200 DVDs each week and inspects them for being good or defective. Using the sample proportion, the quality control inspector tests the null hypothesis \(p \leq .05\) against the alternative hypothesis \(p>.05\), where \(p\) is the proportion of DVDs that are defective. She always uses a \(2.5 \%\) significance level. If the null hypothesis is rejected, the production process is stopped to make any necessary adjustments. A recent sample of 200 DVDs contained 17 defective DVDs. a. Using the \(2.5 \%\) significance level, would you conclude that the production process should be stopped to make necessary adjustments? b. Perform the test of part a using a \(1 \%\) significance level. Is your decision different from the one in part a?

Consider the null hypothesis \(H_{0}: \mu=100\). Suppose that a random sample of 35 observations is taken from this population to perform this test. Using a significance level of \(.01\), show the rejection and nonrejection regions and find the critical value(s) of \(t\) when the alternative hypothesis is as follows. a. \(H_{1}: \mu \neq 100\) b. \(H_{1}: \mu>100\) c. \(H_{1}: \mu<100\)

In 2006, the average number of new single-family homes built per town in the state of Maine was \(14.325\) (www.mainehousing.org). Suppose that a random sample of 42 Maine towns taken in 2009 resulted in an average of \(13.833\) new single-family homes built per town, with a standard deviation of \(4.241\) new single-family homes. Using the \(5 \%\) significance level, can you conclude that the average number of new single-family homes per town built in 2009 in the state of Maine is significantly different from \(14.325\) ? Use both the \(p\) -value and critical-value approaches.

Two years ago, \(75 \%\) of the customers of a bank said that they were satisfied with the services provided by the bank. The manager of the bank wants to know if this percentage of satisfied customers has changed since then. She assigns this responsibility to you. Briefly explain how you would conduct such a test.
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Decision Maths

Read Explanation

Pure Maths

Read Explanation

Statistics

Read Explanation

Discrete Mathematics

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.