/*! 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 32 A hypothesis will be used to tes... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A hypothesis will be used to test that a population mean equals 7 against the alternative that the population mean does not equal 7 with known variance \(\sigma .\) What are the critical values for the test statistic \(Z_{0}\) for the following significance levels? (a) 0.01 (b) 0.05 (c) 0.10

Short Answer

Expert verified
Critical values for (a) 0.01: ±2.576, (b) 0.05: ±1.96, (c) 0.10: ±1.645.

Step by step solution

01

Understanding the Hypothesis Test

We are conducting a two-tailed hypothesis test. The null hypothesis is that the population mean \( \mu = 7 \), and the alternative hypothesis is \( \mu eq 7 \). We want to determine the critical value for the test statistic \( Z_0 \) for different significance levels.
02

Identify the Significance Level

The significance level (\( \alpha \)) represents the probability of rejecting the null hypothesis when it is actually true. We will calculate the critical values for \( \alpha = 0.01 \), \( \alpha = 0.05 \), and \( \alpha = 0.10 \).
03

Z-distribution Critical Values - 0.01 Level

For \( \alpha = 0.01 \) in a two-tailed test, the critical values are found by dividing \( \alpha \) by 2 for each tail: \( \alpha/2 = 0.005 \). Using the standard normal distribution table, the critical z-values are approximately \( \pm 2.576 \).
04

Z-distribution Critical Values - 0.05 Level

For \( \alpha = 0.05 \), divide \( \alpha \) by 2 for each tail: \( \alpha/2 = 0.025 \). From the standard normal distribution table, the critical z-values are approximately \( \pm 1.96 \).
05

Z-distribution Critical Values - 0.10 Level

For \( \alpha = 0.10 \), divide \( \alpha \) by 2 for each tail: \( \alpha/2 = 0.05 \). The critical z-values are approximately \( \pm 1.645 \) from the standard normal distribution table.

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 91Ó°ÊÓ!

Key Concepts

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

Critical Values
In hypothesis testing, critical values play a significant role. They are the boundaries that determine the rejection regions for a test statistic. Critical values help us decide whether to reject or not reject the null hypothesis. When performing a hypothesis test, if the test statistic falls beyond a critical value, we reject the null hypothesis. This concept is crucial because it helps quantify decision-making in statistical testing. Critical values depend on two key factors:
  • The significance level of the test
  • The type of test (one-tailed or two-tailed)
In our example, the two-tailed test has two critical values, one in each tail of the distribution. These values define the cutoff points where the null hypothesis is rejected. It is vital to accurately determine these values to make valid inferences.
Significance Levels
Significance levels, often represented by \( \alpha \), indicate the probability of making a Type I error in hypothesis testing. A Type I error occurs when the null hypothesis is wrongly rejected. It's called a 'level' because it sets a threshold for deciding whether observed data is significantly unusual.Significance levels are pre-determined and play a pivotal role in hypothesis testing. Commonly used levels include:
  • \( \alpha = 0.01 \)
  • \( \alpha = 0.05 \)
  • \( \alpha = 0.10 \)
These choices impact the critical values and consequently the decision to either reject or not reject the null hypothesis. A lower significance level means we require more evidence before rejecting the null hypothesis, reducing the likelihood of a Type I error, but potentially increasing the chance of a Type II error.
Standard Normal Distribution
The standard normal distribution is a special normal distribution with a mean of 0 and a standard deviation of 1. It's represented by the variable \( Z \) and is often used in hypothesis testing because of its simplicity and wide application. In the context of a standard normal distribution, any data point's position can be quantified by a z-score. This z-score tells us how many standard deviations away from the mean a particular value is. When performing a hypothesis test using a standard normal distribution, we often use z-values from the z-table to find the critical values. These z-values correspond to the cumulative probability levels of the significance level split across tails in two-tailed tests. Understanding this distribution is crucial because it simplifies finding critical values in tests where population variance is known.
Two-Tailed Test
A two-tailed test is a statistical test used to determine if a sample is significantly greater or less than a certain range of values. This test is called "two-tailed" because you assess deviations in both directions from the mean. In a two-tailed hypothesis test, the null hypothesis usually states that a parameter equals a certain value, while the alternative hypothesis suggests that the parameter does not equal that value. If we are conducting such a test, the significance level \( \alpha \) is divided into two, placing half in each tail of the distribution. Thus, a critical region is established on both tails, and if the test statistic falls in either region, we reject the null hypothesis. Two-tailed tests are appropriate when deviations in both directions are considered equally important, as it captures more extremes and gives a more balanced view of the parameter's behavior.

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 biotechnology company produces a therapeutic drug whose concentration has a standard deviation of 4 grams per liter. A new method of producing this drug has been proposed, although some additional cost is involved. Management will authorize a change in production technique only if the standard deviation of the concentration in the new process is less than 4 grams per liter. The researchers chose \(n=10\) and obtained the following data in grams per liter. Perform the necessary analysis to determine whether a change in production technique should be implemented. $$\begin{array}{ll}16.628 & 16.630 \\\16.622 & 16.631 \\\16.627 & 16.624 \\\16.623 & 16.622 \\\16.618 & 16.626\end{array}$$

The mean bond strength of a cement product must be at least 1000 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 \(\alpha=0.05 ?\)

A random sample of 500 registered voters in Phoenix is asked if they favor the use of oxygenated fuels yearround to reduce air pollution. If more than 315 voters respond positively, we will conclude that at least \(60 \%\) of the voters favor the use of these fuels. (a) Find the probability of type I error if exactly \(60 \%\) of the voters favor the use of these fuels. (b) What is the type II error probability \(\beta\) if \(75 \%\) of the voters favor this action?

The proportion of adults living in Tempe, Arizona, who are college graduates is estimated to be \(p=0.4 .\) To test this hypothesis, a random sample of 15 Tempe adults is selected. If the number of college graduates is between 4 and 8 , the hypothesis will be accepted; otherwise, you will conclude that \(p \neq 0.4\). (a) Find the type I error probability for this procedure, assuming that \(p=0.4\) (b) Find the probability of committing a type II error if the true proportion is really \(p=0.2\).

A primer paint can be used on aluminum panels. The primer's drying time is an important consideration in the manufacturing process. Twenty panels are selected, and the drying times are as follows: \(1.6,1.3,1.5,1.6,1.7,1.9,1.8,1.6,1.4,\) \(1.8,1.9,1.8,1.7,1.5,1.6,1.4,1.3,1.6,1.5,\) and \(1.8 .\) Is there evidence that the mean drying time of the primer exceeds \(1.5 \mathrm{hr}\) ?

See all solutions

Recommended explanations on Math Textbooks

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.