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 Sci., 1971: 723-736). The sample average toughness was \(\bar{x}=65.6\) for \(m=32\) specimens of the high-purity 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 commercial-purity 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

To determine if the high-purity steel's toughness exceeds commercial-purity steel's by more than 5, set up the hypotheses as follows: \( H_0: \mu_1 - \mu_2 = 5 \) against \( H_a: \mu_1 - \mu_2 > 5 \). Here, \( \mu_1 \) and \( \mu_2 \) are the mean toughness of the high-purity and commercial-purity steel, respectively.

02

Calculate the Test Statistic

The test statistic for the difference in means under known variances is given by \[z = \frac{ (\bar{x} - \bar{y}) - 5 }{\sqrt{ \frac{\sigma_1^2}{m} + \frac{\sigma_2^2}{n} }}\]Substitute the given values: \( \bar{x} = 65.6 \), \( \bar{y} = 59.8 \), \( \sigma_1 = 1.2 \), \( \sigma_2 = 1.1 \), \( m = 32 \), and \( n = 38 \). Calculate:\[z = \frac{ (65.6 - 59.8) - 5 }{\sqrt{ \frac{1.2^2}{32} + \frac{1.1^2}{38} }} = \frac{0.8}{\sqrt{\frac{1.44}{32} + \frac{1.21}{38}}}\]

03

Evaluate the Test Statistic

Calculate the value of the denominator:\[\sqrt{\frac{1.44}{32} + \frac{1.21}{38}} = \sqrt{0.045} \approx 0.212\]Thus, the test statistic becomes:\[z = \frac{0.8}{0.212} \approx 3.77\]

04

Determine the Critical Value and Decision

For a significance level of \( \alpha = 0.001 \), look up the critical \( z \)-value for a one-tailed test, which is approximately 3.09. Since the calculated \( z \) of 3.77 is greater than 3.09, reject the null hypothesis \( H_0 \). The high-purity steel's toughness significantly exceeds commercial-purity steel's by more than 5.

05

Calculate Power of the Test (\beta)

To find \( \beta \) when \( \mu_1 - \mu_2 = 6 \), re-calculate the test statistic assuming the true mean difference is 6:\[z = \frac{ (65.6 - 59.8) - 6 }{0.212} = \frac{-0.2}{0.212} \approx -0.94\]Determine the probability that this \( z \)-value occurs if the alternative is correct, knowing it corresponds to the critical \( z \) = 3.09:\[P(Z < 3.09) - P(Z < -0.94) \approx 0.999 - 0.173 = 0.826\]Hence, \( \beta = 1 - 0.826 = 0.174\).

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Key Concepts

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

Fracture Toughness

Fracture toughness refers to a material's ability to resist crack propagation. It is a crucial property in materials science, especially when assessing the reliability of materials under stress. In practical terms, imagine bending a steel rod; fracture toughness tells us how much force it can withstand before cracking.

High-purity maraging steel often has higher fracture toughness compared to commercial-purity alternatives. This property is vital when considering materials for applications requiring durability and resilience, like aerospace components or high-stress environments.

• Higher fracture toughness means a material can absorb more energy before failure.
• This property is typically measured using tests that apply stress until the material cracks.
• Engineers select materials with suitable toughness to prevent catastrophic failures.

Normal Distribution

A normal distribution, often visualized as a bell curve, describes how values of a variable are distributed. In statistics, it is vital because many traits and measurements, like fracture toughness in our exercise, naturally follow this pattern.

When we say the toughness distributions are normal, it means the majority of toughness values cluster around an average point, with values tapering off symmetrically on either side.

• A normal distribution is defined by its mean (average) and standard deviation (spread).
• It is crucial for hypothesis testing, allowing for standardized calculations and inferences.
• The properties of a normal distribution simplify complex datasets, offering insights into variability and prediction.

Significance Level

In hypothesis testing, the significance level (\( \alpha \)) helps determine the threshold for rejecting our null hypothesis. It is the probability of mistakenly rejecting a true null hypothesis, known as a Type I error.

In our context, with \( \alpha = 0.001 \), there's a tiny 0.1% chance of concluding the high-purity steel has more toughness than commercial steel, when it might not actually be the case. This stringent level ensures high confidence in our decision.

• The smaller the \( \alpha \), the stronger the evidence required to reject the null hypothesis.
• A common choice in other contexts might be 0.05, or 5%.
• Using a smaller value like 0.001 is justified in high-stakes decisions, ensuring reliability and minimizing errors.

Test Statistic

A test statistic is a mathematical tool used to make inferences about a population. It helps us determine whether our observed data are likely under a specific hypothesis.

For testing the difference in toughness between two steel types, we calculated a test statistic using known variances. This statistic, in our case a z-score, measures how far our sample mean difference is from what we expect under the null hypothesis.

• A high absolute value of the test statistic signals stronger evidence against the null hypothesis.
• Test statistics like the z-score allow statisticians to decide whether to reject or accept a hypothesis using standard tables.
• Test statistics are essential for deducing the significance of observed differences or effects.