91Ó°ÊÓ

An experiment was performed to compare the fracture toughness of high-purity \(18 \mathrm{Ni}\) maraging steel with commercial-purity steel of the same type (Corrosion Science, 1971: 723-736). For \(m=32\) specimens, the sample average toughness was \(\bar{x}=65.6\) for the highpurity steel, whereas for \(n=38\) specimens of commercial steel \(\bar{y}=59.8\). Because the high-purity steel is more expensive, its use for a certain application can be justified only if its fracture toughness exceeds that of commercialpurity steel by more than 5 . Suppose that both toughness distributions are normal. a. Assuming that \(\sigma_{1}=1.2\) and \(\sigma_{2}=1.1\), test the relevant hypotheses using \(\alpha=.001\). b. Compute \(\beta\) for the test conducted in part (a) when \(\mu_{1}-\mu_{2}=6\)

Step by step solution

01

State the Hypotheses

We need to test if the fracture toughness of high-purity steel exceeds that of commercial steel by more than 5. Let the null hypothesis be that the difference is 5. Formally, we state:- Null Hypothesis (H_0): \(\mu_1 - \mu_2 = 5\)- Alternative Hypothesis (H_a): \(\mu_1 - \mu_2 > 5\)where \(\mu_1\) and \(\mu_2\) are the population means for high-purity and commercial steel, respectively.

02

Calculate the Test Statistic

We calculate the test statistic using the formula for testing two means with known population standard deviations:\[z = \frac{(\bar{x} - \bar{y}) - (\mu_1 - \mu_2)}{\sqrt{\frac{\sigma_1^2}{m} + \frac{\sigma_2^2}{n}}}\]Substitute the given values: \(\bar{x} = 65.6\), \(\bar{y} = 59.8\), \(\sigma_1 = 1.2\), \(\sigma_2 = 1.1\), \(m = 32\), \(n = 38\), \(\mu_1 - \mu_2 = 5\).\[ z = \frac{(65.6 - 59.8) - 5}{\sqrt{\frac{1.2^2}{32} + \frac{1.1^2}{38}}} = \frac{0.8}{\sqrt{0.045 + 0.031}} = \frac{0.8}{0.270} \approx 2.96 \]

03

Determine the Critical Value and Decision Rule

For a right-tailed test with \(\alpha = 0.001\), check the standard normal distribution table to find the critical z-value. - The critical z-value for \(\alpha = 0.001\) is approximately \(3.090\).Decision Rule:- If \(z > 3.090\), reject H_0.- Else, do not reject H_0.

04

Make a Conclusion for Part (a)

The calculated \(z\)-statistic is \(2.96\), which is less than the critical value of \(3.090\).This result means we fail to reject the null hypothesis at the \(\alpha = 0.001\) level. There is not enough evidence to conclude that the high-purity steel's toughness exceeds that of commercial steel by more than 5.

05

Compute the Power of the Test (Part b)

The power of the test is \(1 - \beta\), where \(\beta\) is the probability of a Type II error. Calculate \(\beta\) for \(\mu_1 - \mu_2 = 6\), using the alternative hypothesis.Substitute into the z formula from Step 2, but replace 5 with 6:\[ z = \frac{(65.6 - 59.8) - 6}{\sqrt{\frac{1.2^2}{32} + \frac{1.1^2}{38}}} = \frac{-0.2}{0.270} = -0.74 \]Find the probability \(P(Z < 3.090 - 0.74)\) using a normal distribution table.- \(Z = 3.090 - (-0.74) = 3.830\)The probability of Z being less than 3.830 is very high (close to 1), so \(\beta\) is very small.Therefore, the power is close to \(1\).

06

Finalize \(\beta\) Calculation and Power of Test

Continue the power calculation based on the previous step:Since \(P(Z < 3.830)\) is nearly 1 (from normal distribution tables), \(\beta\) (the probability of Type II error) is approximately 0.Thus, the power of the test is approximately \(1 - \beta = 1\), meaning if \(\mu_1 - \mu_2 = 6\), the test almost certainly detects this difference.

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Key Concepts

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

Statistical Power

Statistical power is an essential component of hypothesis testing, referring to the probability that a test will correctly reject a false null hypothesis. In simpler terms, it helps us understand how likely we are to detect an actual effect or difference when it truly exists. High statistical power, typically close to 1, indicates a strong ability to detect true differences.

In our example, we calculated the power of the test to find if the difference in fracture toughness between high-purity and commercial-purity steel was significant. We found that the statistical power was nearly 1 when the true difference was 6, meaning the test is very effective in this scenario.

• A high statistical power reduces the risk of a Type II error.
• Statistical power depends on several factors, including sample size and effect size.

Type II Error

A Type II error happens when we fail to reject a null hypothesis that is actually false. It's like saying there's no difference when, in reality, there is. The probability of making a Type II error is denoted by \( \beta \).

In the context of our exercise, we're concerned about failing to conclude that the high-purity steel has a significantly greater toughness than the commercial steel, when it truly does. Controlling this error is crucial because making a Type II error might lead us to overlook an important finding.

• Type II errors occur more frequently with small sample sizes or underpowered tests.
• The complement of \( \beta \) (1 - \( \beta \)), called the power of the test, helps to gauge how effectively you're avoiding a Type II error.

Z-Test

The \( z \)-test is a statistical test used to determine if there is a significant difference between sample means compared to a predicted value. It's applicable when we know the population standard deviation.

In our study, we used a \( z \)-test to compare the fracture toughness of two types of steel. We calculated a \( z \)-statistic to measure how far the observed difference is from the hypothesized value under the null hypothesis.

• The formula for the \( z \)-test statistic is given by \( z = \frac{(\bar{x} - \bar{y}) - (\mu_1 - \mu_2)}{\sqrt{\frac{\sigma_1^2}{m} + \frac{\sigma_2^2}{n}}} \).
• A higher absolute \( z \)-value indicates a more significant difference.

Critical Value

Critical values are thresholds that determine whether to reject the null hypothesis in a hypothesis test. They are defined based on the significance level (\( \alpha \)), which is the probability of committing a Type I error.

In our example, we set \( \alpha = 0.001 \), indicating a high level of confidence required to reject the null hypothesis. The critical value for this \( \alpha \) in a right-tailed \( z \)-test is 3.090. The calculated \( z \)-statistic of 2.96, which is less than the critical value, led us not to reject the null hypothesis.

• A \( z \)-value greater than the critical value indicates significant results at the given \( \alpha \).
• Critical values vary depending on whether the test is one-tailed or two-tailed.