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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

Mong Corporation makes auto batteries. The company claims that \(80 \%\) of its LL.70 batteries are good for 70 months or longer. A consumer agency wanted to check if this claim is true. The agency took a random sample of 40 such batteries and found that \(75 \%\) of them were good for 70 months or longer. a. Using the \(1 \%\) significance level, can you conclude that the company's claim is false? b. What will your decision be in part a if the probability of making a Type I error is zero? Explain.

In Las Vegas and Atlantic City, New Jersey, tests are performed often on the various gaming devices used in casinos. For example, dice are often tested to determine if they are balanced. Suppose you are assigned the task of testing a die, using a two-tailed test to make sure that the probability of a 2 -spot is \(1 / 6 .\) Using the \(5 \%\) significance level, determine how many 2 -spots you would have to obtain to reject the null hypothesis when your sample size is \(\begin{array}{lll}\text { a. } 120 & \text { b. } 1200 & \text { c. } 12,000\end{array}\) Calculate the value of \(\hat{p}\) for each of these three cases. What can you say about the relationship between (1) the difference between \(\hat{p}\) and \(1 / 6\) that is necessary to reject the null hypothesis and (2) the sample size as it gets larger?

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 \(\alpha=.025\).

A company claims that the mean net weight of the contents of its All Taste cereal boxes is at least 18 ounces. Suppose you want to test whether or not the claim of the company is true. Explain briefly how you would conduct this test using a large sample. Assume that \(\sigma=.25\) ounce.

The customers at a bank complained about long lines and the time they had to spend waiting for service. It is known that the customers at this bank had to wait 8 minutes, on average, before being served. The management made some changes to reduce the waiting time for its customers. A sample of 60 customers taken after these changes were made produced a mean waiting time of \(7.5\) minutes with a standard deviation of \(2.1\) minutes. Using this sample mean, the bank manager displayed a huge banner inside the bank mentioning that the mean waiting time for customers has been reduced by new changes. Do you think the bank manager's claim is justifiable? Use the \(2.5 \%\) significance level to answer this question. Use both approaches.
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