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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.

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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.

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Most popular questions from this chapter

A study claims that all adults spend an average of 14 hours or more on chores during a weekend. A researcher wanted to check if this claim is true. A random sample of 200 adults taken by this researcher showed that these adults spend an average of \(14.65\) hours on chores during a weekend. The population standard deviation is known to be \(3.0\) hours. a. Find the \(p\) -value for the hypothesis test with the alternative hypothesis that all adults spend more than 14 hours on chores during a weekend. Will you reject the null hypothesis at \(\alpha=.01\) ? b. Test the hypothesis of part a using the critical-value approach and \(\alpha=.01\).

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 \(\alpha=.05\) ?

Shulman Steel Corporation makes bearings that are supplied to other companies. One of the machines makes bearings that are supposed to have a diamcter of 4 inches. The bearings that have a diameter of either more or less than 4 inches are considered defective and are discarded. When working properly, the machine does not produce more than \(7 \%\) of bearings that are defective. The quality control inspector selects a sample of 200 bearings each week and inspects them for the size of their diameters. Using the sample proportion, the quality control inspector tests the null hypothesis \(p \leq .07\) against the alternative hypothesis \(p \geq\) 07, where \(p\) is the proportion of bearings that are defective. He always uses a \(2 \%\) significance level. If the null hypothesis is rejected, the machine is stopped to make any necessary adjustments. One sample of 200 bearings taken recently contained 22 defective bearings. a. Using the \(2 \%\) significance level, will you conclude that the machine 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?

A mail-order company claims that at least \(60 \%\) of all orders are mailed within 48 hours. From time to time the quality control department at the company checks if this promise is fulfilled. Recently the quality control department at this company took a sample of 400 orders and found that 208 of them were mailed within 48 hours of the placement of the orders. a. Testing at the \(1 \%\) significance level, can you conclude that the company's claim is true? b. What will your decision be in part a if the probability of making a Type I error is zero? Explain. c. Make the test of part a using the \(p\) -value approach and \(\alpha=.01\).

What are the four possible outcomes for a test of hypothesis? Show these outcomes by writing a table. Briefly describe the Type I and Type II errors.
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