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Chapter 16: Problem 641

We desire to perform an experiment to determine whether surface finish has an effect on the endurance limit of steel. There exists a theory that polishing increases the average endurance limit. If in our experiment we wish to detect that polishing fails to have an effect with a probability of \(0.99\) \((\alpha=.01)\) and we also wish to detect a change of 7500 units by a probability of at least \(0.9\), then if it is known that the standard deviation of the endurance limit of the steel is 4000 units, what sample size would be needed to carry out this experiment?

Short Answer

Expert verified
A sample size of 11 is needed to carry out this experiment with a desired significance level of \(\alpha=0.01\), and a desired power of \(1-\beta=0.9\), to detect a change of 7500 units in the endurance limit of steel with a standard deviation of 4000 units.

Step by step solution

01

Define the hypotheses

We will perform a hypothesis test to check if polishing has an effect on the endurance limit of steel. The null hypothesis (H0) is that polishing has no effect, and the alternative hypothesis (H1) is that polishing has an effect (increases the endurance limit). We can define: - H0: \(\mu_1 = \mu_2\) - H1: \(\mu_1 \neq \mu_2\)
02

Use the z-score and power calculations to find sample size

We know that the significance level (\(\alpha\)) and the desired power (\(1-\beta\)) are \(0.01\) and \(0.9\), respectively. We can find the corresponding z-scores using the standard normal distribution: - \(z_\alpha = z_{0.01} = 2.576\) (the z-score corresponding to a 0.01 significance level) - \(z_\beta = z_{0.1} = 1.282\) (the z-score corresponding to 0.1, because \(1-\beta = 0.9\)) Next, we'll use the formula for sample size in power analysis with a difference (\(\Delta\)) to be detected and \(\sigma\) as the standard deviation: $$n = \frac{(\sigma(z_\alpha + z_\beta))^2}{\Delta^2}$$
03

Calculate the sample size

Using the values given, we have: - \(\Delta = 7500\) (the difference we want to detect) - \(\sigma = 4000\) (the standard deviation) - \(z_\alpha = 2.576\) - \(z_\beta = 1.282\) Plug these values into the sample size formula: $$n = \frac{(4000(2.576 + 1.282))^2}{7500^2}$$
04

Calculate and report the result

Calculate the value for n: $$n = \frac{(4000(3.858))^2}{7500^2} \approx 10.656$$ Since we cannot have a fraction of a sample, we need to round up to the nearest whole number: $$n = 11$$ Thus, a sample size of 11 is needed to carry out this experiment.

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Key Concepts

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

Sample Size Determination
Determining the right sample size is crucial for ensuring that an experiment or study provides reliable results. A sample that is too small might not adequately capture the effects being studied, while a too large one may waste resources.

To calculate the ideal sample size, certain factors must be considered:
  • Significance Level (\( \alpha \): The probability of rejecting the null hypothesis when it is true. Commonly set at 0.05, it was 0.01 in this case, suggesting a stricter requirement for significant findings.
  • Power (\(1-\beta\)): The probability of correctly rejecting a false null hypothesis. In this experiment, the desired power was 0.9, indicating a high likelihood of detecting a genuine effect.
  • Expected Effect Size: The magnitude of the difference to be detected, here set at 7500 units.
  • Standard Deviation (\(\sigma\)): Variation within the data, known as 4000 units in this study.
The formula used: \[ n = \frac{(\sigma(z_\alpha + z_\beta))^2}{\Delta^2} \]provides a methodical way to compute sample size, balancing practical considerations with statistical theory.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. It tells us how much the values in a data set differ from the mean of the data set.

In the context of this experiment on the endurance limit of steel, the standard deviation of 4000 units indicates how much individual endurance limits, when tested, are expected to vary around the average endurance limit.

Knowing the standard deviation is essential for understanding:
  • The spread of values in the experiment's results.
  • How precise the measurements are and their reliability.
  • The ability to calculate sample size, as the formula for sample size directly involves standard deviation.
A larger standard deviation suggests greater variability, possibly requiring a larger sample to accurately detect any true effects or differences in the study.
Significance Level
The significance level determines the threshold at which you reject the null hypothesis. Also known as alpha \((\alpha)\), it represents the probability of making a type I error 鈥� incorrectly rejecting the null hypothesis when it is in fact true.

For this study, a significance level of 0.01 was chosen. A lower \(\alpha\) value like this indicates that strong evidence is needed to support a rejection of the null hypothesis, reducing the chances of a false positive finding.

Key implications of setting the significance level include:
  • Risk of Type I Error: Lower \(\alpha\) minimizes this risk, tolerating less chance of a mistaken rejection of the null hypothesis.
  • Statistical Rigour: A more stringent significance level suggests a need for more compelling evidence to declare a result statistically significant.
Understanding the significance level is crucial for ensuring experimental findings are robust and credible.
Power of a Test
The power of a test indicates how effectively an experiment can detect an effect when one truly exists. It is represented as \(1-\beta\), where beta \((\beta)\) is the probability of making a type II error - failing to reject the null hypothesis when it is false.

In this analysis, the desired power was 0.9, meaning there is a 90% chance of detecting a true effect or difference if it exists. High power is essential for:
  • Achieving Reliable Results: Higher power increases confidence in the experiment鈥檚 findings.
  • Minimizing Type II Errors: Reduces the risk of overlooking a true effect.
  • Determining Sample Size: High power typically requires a larger sample size to substantiate the test鈥檚 findings.
In conclusion, having high test power is vital for ensuring the robustness and reliability of the experimental outcomes and for strategically allocating resources toward sample collection.

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Most popular questions from this chapter

A college dean wants to determine if the students entering the college in a given year have higher IQ's than the students entering the same college in the previous year. The IQ's of the college entrants in the two years are known to be normally distributed and to have equal variances. A random sample of four students from this year's entering class gives IQ scores of \(110,113,116\), and 117. Another random sample of four students from last year's entering class gives IQ scores of \(109,111,112\), and 112 . At the \(5 \%\) level of significance, can we conclude that this year's students have a higher IQ than last years?

For the following samples of data, compute \(\mathrm{t}\) and determine whether \(\mu_{1}\) is significantly less than \(\mu_{2}\). For your test use a level of significance of \(10 .\) Sample \(1: \mathrm{n}=10, \underline{\mathrm{X}}=10.0\) \(\mathrm{S}=5.2\); sample \(2: \mathrm{n}=10, \underline{\mathrm{X}}=13.3, \mathrm{~S}=5.7\)

Consider the probability distribution function $$ \mathrm{f}(\mathrm{x} ; \theta)=(1 / \theta) \mathrm{e}^{-\mathrm{x} / \theta} \quad 0<\mathrm{x}<\infty $$ $$ =0 $$ elsewhere. It is desired to test the hypothesis \(\mathrm{H}_{0}: \theta=2\) against alternate hypothesis \(\mathrm{H}_{1}: \theta>2\). Suppose a random sample \(\mathrm{X}_{1}, \mathrm{X}_{2}\) is used and the critical region is \(\mathrm{X}_{1}+\mathrm{X}_{2} \geq 9.5\) Calculate an expression for the power function, \(\mathrm{K}\left(\theta_{1}\right)\), for all \(\theta_{1}>2\), and specifically for \(\theta_{1}=4\).

From appropriately selected samples, two sets of IQ scores are obtained. For group \(1, \underline{\mathrm{X}}=104, \mathrm{~S}=10\), and \(\mathrm{n}=16\); for group \(2, \underline{\mathrm{X}}=112, \mathrm{~S}=8\), and \(\mathrm{n}=14\). At the \(5 \%\) significance level is there a significant difference between the 2 groups?

An official of a trade union reports that the mean yearly wage is \(\$ 8,000\). A random sample of 100 employees in the union produced a mean of \(\$ 7,875\) with a standard deviation of \(\$ 1,000\). Test the null hypothesis at the 05 level of significance that the mean wage is \(\$ 8,000\) against the alternate hypothesis that the wage is greater than or less than \(\$ 8000\)
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