/*! 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 26 A random sample of 80 observatio... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 26

A random sample of 80 observations produced a sample mean of \(86.50\). Find the critical and observed values of \(z\) for each of the following tests of hypothesis using \(\alpha=.10 .\) The population standard deviation is known to be \(7.20\). a. \(H_{0}: \mu=91\) versus \(H_{1}: \mu \neq 91\) b. \(H_{0}: \mu=91 \quad\) versus \(\quad H_{1}: \mu<91\)

Short Answer

Expert verified
The critical z values are \(卤1.645\) for the two-tailed test and \(-1.282\) for the left-tailed test. The observed z value for both tests is \(-4.73\).

Step by step solution

01

- Identify Hypotheses and Significance Level

For both tests, our null hypothesis \(H_{0}\) is that the population mean \(\mu\) is equal to 91. For the first test, the alternative hypothesis \(H_{1}\) is that \(\mu\) is not equal to 91, making this a two-tailed test. For the second test, \(H_{1}\) is that \(\mu\) is less than 91, making it a left-tailed test. The significance level \(\alpha\) is the same for both tests, given as 0.10.
02

- Find the Critical Z Value

The critical z value is the z score that corresponds to the given level of confidence. For a two-tailed test, the significance level is divided between the two tails. For an \(\alpha\) of 0.10, there will be 0.05 in each tail, yielding critical z values of \(卤1.645\). In a left-tailed test, the entire \(\alpha\) of 0.10 is in the left tail, yielding a critical z value of \(-1.282\).
03

- Calculate the Observed Z Value

We calculate the z score using the formula \(Z = \frac{(\bar{X} - \mu)}{(\sigma/\sqrt{n})}\), where \(\bar{X}\) is the sample mean, \(\mu\) is the population mean, \(\sigma\) is the population standard deviation, and \(n\) is the sample size. Substituting the provided data, we get: \(Z = \frac{(86.50 - 91)} {(7.20/\sqrt{80})} = -4.73\). This is the observed z value for both tests.

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.

Two-Tailed Test
When you are exploring whether a sample mean differs from a population mean, you might conduct a two-tailed test. In these types of tests, the alternative hypothesis (\(H_1\)) suggests that the sample mean can be either greater than or less than the population mean. Thus, you are investigating both directions.
In practice:
  • The rejection area is found on both ends of the normal distribution.
  • The total significance level is split between the two tails.
  • This type is useful when deviations in either direction are important.
For example, if you are examining a hypothesis where the null is \(\mu = 91\) and you have \(\mu eq 91\) as your alternative, a result significantly higher or lower from 91 will result in rejecting the null hypothesis.
Left-Tailed Test
In a left-tailed test, the focus shifts. Here, the alternative hypothesis (\(H_1\)) specifies that the sample mean is less than the population mean. This type zeroes in on one direction.
With left-tailed tests:
  • The rejection region exists solely on the left end of the distribution curve.
  • The entire significance level is dedicated to this one tail.
  • This method is optimal when you're only interested in decrease or negative deviation from the population mean.
Consider a scenario where your null hypothesis is \(\mu = 91\) and the alternative is \(\mu < 91\). Your approach is concentrated on detecting decreases.
Critical Z Value
The critical \(z\) value represents a threshold. It is the boundary beyond which the null hypothesis is rejected in favor of the alternative hypothesis. It's determined based on the significance level (\(\alpha\)) and whether the test is one-tailed or two-tailed.
To determine:
  • For a two-tailed test with \(\alpha=0.10\), the critical \(z\) values are \(卤1.645\), as the \(\alpha\) is split to \(0.05\) in each tail.
  • For a left-tailed test, with the same \(\alpha\), the critical \(z\) value is \(-1.282\), focusing entirely on one tail.
These values guide the decision-making process in hypothesis testing.
Observed Z Value
Let's calculate the observed \(z\) value. This is your computed value from sample data. It determines where your sample mean falls within the normal distribution relative to the null hypothesis.
You find it using the formula:\[Z = \frac{(\bar{X} - \mu)}{(\sigma/\sqrt{n})}\]Where:
  • \(\bar{X}\) = sample mean (e.g., 86.50)
  • \(\mu\) = claimed population mean
  • \(\sigma\) = population standard deviation
  • \(n\) = sample size
Suppose \(\bar{X}=86.50\), \(\mu=91\), \(\sigma=7.20\), and \(n=80\), the observed \(z\) becomes \(-4.73\). This shows a noticeable deviation from the null hypothesis.
Significance Level
The significance level, denoted as \(\alpha\), is a probability threshold defining how extreme observed results must be to reject the null hypothesis. It's key in hypothesis testing and reflects the risk of incorrectly rejecting a true null hypothesis.
Keep in mind:
  • A lower \(\alpha\) implies a stricter test, reducing false positives but potentially missing true effects.
  • In our discussion, \(\alpha=0.10\), a moderate level indicating a 10% risk threshold for our two-tailed and left-tailed hypothesis tests.
  • Significance levels guide the location of the critical \(z\) values.
Choosing an \(\alpha\) balances between Type I errors (false positives) and Type II errors (false negatives) in your testing.

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

Records in a three-county area show that in the last few years, Girl Scouts sell an average of \(47.93\) boxes of cookies per year, with a population standard deviation of \(8.45\) boxes per year. Fifty randomly selected Girl Scouts from the region sold an average of \(46.54\) boxes this year. Scout leaders are concerned that the demand for Girl Scout cookies may have decreased. a. Test at the \(10 \%\) significance level whether the average number of boxes of cookies sold by all Girl Scouts in the three-county area is lower than the historical average. b. What will your decision be in part a if the probability of a Type I error is zero? Explain.

Consider the following null and alternative hypotheses: $$ H_{0}: \mu=120 \text { versus } H_{1}: \mu>120 $$ A random sample of 81 observations taken from this population produced a sample mean of \(123.5 .\) The population standard deviation is known to be 15 . a. If this test is made at the \(2.5 \%\) significance level, would you reject the null hypothesis? Use the critical-value approach. b. What is the probability of making a Type I error in part a? c. Calculate the \(p\) -value for the test. Based on this \(p\) -value, would you reject the null hypothesis if \(\alpha=.01 ?\) What if \(\alpha=.05\) ?

The past records of a supermarket show that its customers spend an average of \(\$ 95\) per visit at this store. Recently the management of the store initiated a promotional campaign according to which each customer receives points based on the total money spent at the store, and these points can be used to buy products at the store. The management expects that as a result of this campaign, the customers should be encouraged to spend more money at the store. To check whether this is true, the manager of the store took a sample of 14 customers who visited the store. The following data give the money (in dollars) spent by these customers at this supermarket during their visits. \(\begin{array}{rrrrrrr}109.15 & 136.01 & 107.02 & 116.15 & 101.53 & 109.29 & 110.79 \\ 94.83 & 100.91 & 97.94 & 104.30 & 83.54 & 67.59 & 120.44\end{array}\) Assume that the money spent by all customers at this supermarket has a normal distribution. Using the \(5 \%\) significance level, can you conclude that the mean amount of money spent by all customers at this supermarket after the campaign was started is more than \(\$ 95 ?\) (Hint: First calculate the sample mean and the sample standard deviation for these data using the formulas learned in Sections \(3.1 .1\) and \(3.2 .2\) of Chapter \(3 .\) Then make the test of hypothesis about \(\mu .\) )

Explain how the tails of a test depend on the sign in the alternative hypothesis. Describe the signs in the null and alternative hypotheses for a two-tailed, a left-tailed, and a right-tailed test, respectively.

A telephone company claims that the mean duration of all long-distance phone calls made by its residential customers is 10 minutes. A random sample of 100 long-distance calls made by its residential customers taken from the records of this company showed that the mean duration of calls for this sample is \(9.20\) minutes. The population standard deviation is known to be \(3.80\) minutes. a. Find the \(p\) -value for the test that the mean duration of all long- distance calls made by residential customers is different from 10 minutes. If \(\alpha=.02\), based on this \(p\) -value, would you reject the null hypothesis? Explain. What if \(\alpha=.05\) ? b. Test the hypothesis of part a using the critical-value approach and \(\alpha=.02\). Does your conclusion change if \(\alpha=.05 ?\)
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Applied Mathematics

Read Explanation

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

Read Explanation

Geometry

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.