91Ó°ÊÓ

Reconsider the paint-drying problem discussed in Example 9.2. The hypotheses were \(H_{0}: \mu=75\) versus \(H_{\mathrm{a}}: \mu<75\), with \(\sigma\) assumed to have value 9.0. Consider the alternative value \(\mu=74\), which in the context of the problem would presumably not be a practically significant departure from \(H_{0}\). a. For a level .01 test, compute \(\beta\) at this alternative for sample sizes \(n=100,900\), and 2500 . b. If the observed value of \(\bar{X}\) is \(\bar{x}=74\), what can you say about the resulting \(P\)-value when \(n=2500 ?\) Is the data statistically significant at any of the standard values of \(\alpha\) ? c. Would you really want to use a sample size of 2500 along with a level \(.01\) test (disregarding the cost of such an experiment)? Explain.

Step by step solution

01

Understanding Hypotheses and Parameters

We have the null hypothesis \(H_0: \mu = 75\) and the alternative hypothesis \(H_a: \mu < 75\). Given \(\sigma = 9.0\), we need to calculate \(\beta\), which is the probability of Type II error, for different sample sizes \(n = 100, 900,\) and \(2500\). The alternative mean is assumed to be \(\mu = 74\).

02

Calculate Critical Value for \(\alpha = 0.01\)

For a level \(0.01\) test, calculate the critical value (\(z_{\alpha}\)) from the standard normal distribution for \(\alpha = 0.01\). This value corresponds to the lower tail since our hypothesis is left-tailed. It is approximately \(z_{\alpha} = -2.33\).

03

Calculate Type II Error \(\beta\) for \(n = 100\)

For \(n = 100\), the standard error is \(\sigma_{\bar{X}} = \frac{9}{\sqrt{100}} = 0.9\). Calculate \(z\) for \(\mu = 74\):\[ z = \frac{74 - 75}{0.9} = -1.11 \]Find \(\beta\) using the normal distribution from \(z = -1.11\) to the critical value \(z = -2.33\):\[ \beta_{100} = P(Z > -2.33) - P(Z > -1.11) \approx 0.1335 \]

04

Calculate Type II Error \(\beta\) for \(n = 900\)

For \(n = 900\), the standard error is \(\sigma_{\bar{X}} = \frac{9}{\sqrt{900}} = 0.3\). Calculate \(z\) for \(\mu = 74\):\[ z = \frac{74 - 75}{0.3} = -3.33 \]The corresponding \(\beta\) is:\[ \beta_{900} = P(Z > -2.33) - P(Z > -3.33) \approx 0.0093 \]

05

Calculate Type II Error \(\beta\) for \(n = 2500\)

For \(n = 2500\), the standard error is \(\sigma_{\bar{X}} = \frac{9}{\sqrt{2500}} = 0.18\). Calculate \(z\) for \(\mu = 74\):\[ z = \frac{74 - 75}{0.18} = -5.56 \]The corresponding \(\beta\) is essentially \(0\) because \(P(Z < -5.56)\) is nearly zero.

06

Determine P-value for \(n = 2500\) with \(\bar{x} = 74\)

Calculate \(z\) value for observed \(\bar{x} = 74\):\[ z = \frac{74 - 75}{0.18} = -5.56 \]The P-value is the probability \(P(Z < -5.56)\), which is very small (less than 0.00001), showing high statistical significance.

07

Conclusion about Sample Size 2500 and Level 0.01

A sample size of 2500 makes the test extremely sensitive, with an essentially zero \(\beta\), indicating that very small departures from \(\mu = 75\) can be detected. However, this might not be practical if the goal is to detect only meaningful changes. The extremely large sample also magnifies the statistical significance, potentially highlighting results that aren't practically significant.

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Key Concepts

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

Type I and Type II Errors

When performing hypothesis testing, understanding Type I and Type II errors is crucial.
- **Type I Error** occurs when the null hypothesis (\(H_0\)) is true, but mistakenly rejected. This is the error we control with the significance level, \( \alpha \), such as 0.01. Essentially, \( \alpha \) is the probability of making a Type I error.
- **Type II Error** happens when the null hypothesis is false, but we fail to reject it. The probability of this error is denoted as \( \beta \).
In the paint-drying example:

• We calculate \( \beta \) based on different sample sizes.
  
• As the sample size increases, \( \beta \) generally decreases. This means less chance of not detecting when \( \mu \) is truly not equal to 75.
  

Knowing both types of errors helps in designing tests that provide a balance between risk of false alarm (Type I) and missing a significant change (Type II).

Normal Distribution

The normal distribution is a key concept in hypothesis testing. It's a bell-shaped distribution used to describe the distribution of sample means, especially due to the Central Limit Theorem.
In our exercise, the population standard deviation \( \sigma = 9 \) allows us to calculate the standard error:

• The standard error (\(\sigma_{\bar{X}}\)) of the sample means helps determine how much the sample mean deviates from the hypothesized population mean.
  
• Using the standard error with different sample sizes, we calculate the corresponding \( z \) scores which reflect the distance of the sample mean from the hypothesized mean, in units of standard error.
  

Using the properties of the normal distribution, we find critical values and calculate probabilities of committing Type I and Type II errors, assisting the determination of the strength of our statistical test.

Significance Level

The **significance level**, \( \alpha \), is a threshold set by the researcher to determine the cutoff for rejecting the null hypothesis. In our problem, a significance level of 0.01 is used.
Here's what it signifies:

• It represents a 1% risk of concluding a result is significant when it actually isn't.
  
• Lower \( \alpha \) (e.g., 0.01 instead of 0.05) means stricter criteria for significance, reducing Type I error risk.
  

In the paint-drying exercise, determining the critical value associated with \( \alpha = 0.01 \) allows identifying how extreme a test statistic must be to reject \(H_0\). The test is extremely sensitive with a very small \(\beta\), therefore, a sample size of 2500 leads us to a very low \(P\)-value, showing statistical significance even for small deviations from the null hypothesis. This highlights the importance of choosing \(\alpha\) appropriately, considering practical significance beyond just statistical calculations.