91Ó°ÊÓ

Two alloys \(A\) and \(B\) are being used to manufacture a certain steel product. An experiment needs to be designed to compare the two in terms of maximum load capacity in tons. This is the maximum that can be tolerated without breaking. It is known that the two standard deviations in load capacity are equal at 5 tons each. An experiment is conducted in which 30 specimens of each alloy (A and B) are tested and the results gave $$\bar{x}_{A}=49.5, \quad \bar{x}_{B}=45.5 ; \quad \bar{X} A-\bar{x}_{B}=4$$ The manufacturers of alloy \(A\) are convinced that this evidence shows conclusively that \(\mu_{A}>\mu_{B}\) and strongly supports their alloy. Manufacturers of alloy \(B\) claim that the experiment could easily have given \(x_{A}-x_{B}=\) 4 even if the two population means arc: equal. In other words, "tilings are inconclusive!" (a) Make an argument that manufacturers of alloy \(B\) are wrong. Do it by computing $$P\left(\bar{X}_{A_{-}} X_{B}^{-}>4 \mid \mu_{A}=\mu_{B}\right).$$ (b) Do you think these data strongly support alloy \(A\) ?

Step by step solution

01

Stating the Hypotheses

The null hypothesis will be \(\mu_{A} = \mu_{B}\), which means the mean load capacities of both alloys are equal. The alternate hypothesis is \(\mu_{A} > \mu_{B}\), meaning that alloy A has a greater load capacity than alloy B.

02

Computing the Standard Error

The next step is to calculate the standard error. The formula is \(\sqrt{\sigma_{1}^{2}/n_{1} + \sigma_{2}^{2}/n_{2}}\), simplified to \(\sqrt{2 * (\sigma^{2}/n)} = \sqrt{2 * (5^{2}/30)}\) because \(\sigma_{1} = \sigma_{2} = 5\) tons and \(n_{1} = n_{2} = 30\) specimens. This approximates to 1.82574.

03

Calculating the Test Statistic

The test statistic t is given by \((\bar{x}_{1} - \bar{x}_{2}) / SE\), where \(\bar{x}_{1} - \bar{x}_{2}\) is the difference between the two sample means and SE is the standard error. By substituting the given values, the result is \(4 / 1.82574\) which approximates to 2.19178.

04

Calculating the P-value

The aim now is to compute the P-value which can be found using the statistical tables for the t-distribution for a degree of freedom \((n_{1} + n_{2} - 2 = 30 + 30 - 2 = 58)\) and the computed t-value, resulting in a P-value that is less than 0.05.

05

Conclusion

Since the P-value is less than the common significance level of 0.05, the null hypothesis is rejected. This means that there is strong evidence that alloy A is stronger.

06

Interpretation

The data strongly supports alloy A as the manufacturers claimed. Alloy B manufacturers are wrong based on this specific experiment's data and statistical inference. However, final conclusions always need replication and additional evidence.

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Key Concepts

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

Standard Error

In hypothesis testing, understanding the standard error (SE) is crucial because it helps determine the precision of a sample mean. The standard error is the standard deviation of the sampling distribution of a statistic, usually the mean. It's computed as the following:

• The formula: \( \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}} \)
• In our problem, the standard deviations are equal and the sample sizes are the same, simplifying the formula to \( \sqrt{2 \times \left(\frac{\sigma^2}{n}\right)} \)

This gives us a standard error of approximately 1.82574 tons. Why does this matter? Because it allows us to understand how much the sample means we have could vary if we repeated the experiment multiple times. A smaller SE suggests more reliable sample means. Therefore, it quantifies the variability and reliability of our experimental results.

t-distribution

The t-distribution is a type of probability distribution that is symmetric and bell-shaped, similar to the normal distribution but with heavier tails. This feature accounts for the extra variability that comes from estimating the population standard deviation using the sample standard deviation.

• The t-distribution is especially useful when dealing with small sample sizes.
• In our case, we use it because we're working with the means from samples of size 30.

To calculate the test statistic in hypothesis testing, the formula used is:\[t = \frac{\bar{x}_1 - \bar{x}_2}{SE}\]Where \(\bar{x}_1 - \bar{x}_2\) is the difference between sample means and SE is the standard error calculated earlier. With our data, this results in a t-value of approximately 2.19178. This t-value is then used to find the P-value, which helps us determine the significance of our test results.

P-value

The P-value is a crucial component of hypothesis testing. It measures the probability of obtaining a test result at least as extreme as the one observed, under the assumption that the null hypothesis is true. Simply put, it tells us how likely our data would occur by random chance alone.

• A smaller P-value indicates stronger evidence against the null hypothesis.
• If the P-value is less than a predetermined significance level (commonly 0.05), the results are considered statistically significant.

In our exercise, after calculating the test statistic and looking it up in the t-distribution table for 58 degrees of freedom, we find a P-value less than 0.05. This small P-value allows us to reject the null hypothesis, providing strong evidence that alloy A has a greater load capacity compared to alloy B.

Sample Means

Sample means play a significant role in hypothesis testing. They represent the average values obtained from sample data and serve as estimates of the population mean.

• They are calculated by summing all sample observations and dividing by the number of observations.
• The variability of sample means is less than that of individual observations, leading to a more stable estimate of the population mean.

In our test, the means obtained from each sample were \(\bar{x}_{A} = 49.5\) and \(\bar{x}_{B} = 45.5\). The difference between these sample means, \(\bar{x}_{A} - \bar{x}_{B} = 4\), suggests that on average, the specimens made from alloy A can withstand 4 more tons than those made from alloy B. Understanding this helps us form hypotheses about population parameters and evaluate the results in the context of the experiment.