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Chapter 9: Problem 9

Two different types of alloy, A and B, have been used to manufacture experimental specimens of a small tension link to be used in a certain engineering application. The ultimate strength (ksi) of each specimen was determined, and the results are summarized in the accompanying frequency distribution. \begin{tabular}{lrr} & \(\mathbf{A}\) & \(\mathbf{B}\) \\ \hline \(26-<30\) & 6 & 4 \\ \(30-<34\) & 12 & 9 \\ \(34-<38\) & 15 & 19 \\ \(38-<42\) & 7 & 10 \\ & \(m=40\) & \(m=42\) \\ \hline \end{tabular} Compute a \(95 \%\) CI for the difference between the true proportions of all specimens of alloys \(\mathrm{A}\) and \(\mathrm{B}\) that have an ultimate strength of at least \(34 \mathrm{ksi}\).

Short Answer

Expert verified
The 95% confidence interval for the difference is \(-0.3431 \text{ to } 0.0621\).

Step by step solution

01

Determine the relevant data for each alloy

For alloy A, we need to calculate the number of specimens with ultimate strength of at least 34 ksi. These are from the intervals 34-<38 and 38-<42. Thus, the number for A is 15 + 7 = 22. For alloy B, the relevant intervals are the same, giving us 19 + 10 = 29 specimens.
02

Calculate the proportions for each alloy

The proportion of specimens with at least 34 ksi strength for alloy A is \( p_A = \frac{22}{40} = 0.55 \). For alloy B, the proportion is \( p_B = \frac{29}{42} \approx 0.6905 \).
03

Calculate the standard error of the difference in proportions

The standard error (SE) for the difference between two proportions is given by \( SE = \sqrt{\frac{p_A(1-p_A)}{m_A} + \frac{p_B(1-p_B)}{m_B}} \). Substituting the values, we have:\( SE = \sqrt{\frac{0.55 \times 0.45}{40} + \frac{0.6905 \times 0.3095}{42}} \approx \sqrt{0.0061875 + 0.0050906} \approx 0.1036 \).
04

Determine the critical value for the confidence interval

For a 95% confidence interval, the critical value (z-score) is approximately 1.96 (from a standard normal distribution table).
05

Calculate the confidence interval

The 95% confidence interval is calculated as the difference in proportions \((p_A - p_B)\), plus and minus the product of the critical value and the standard error: \[ (p_A - p_B) \pm 1.96 \times SE \]Substituting the values, we get:\((0.55 - 0.6905) \pm 1.96 \times 0.1036 \approx -0.1405 \pm 0.2026 \),which is \(-0.3431 \text{ to } 0.0621\).

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Key Concepts

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

Proportions
Proportions help us understand the part of a whole we're examining. In statistics, a proportion is essentially a fraction of a total amount that meets specific criteria. In our problem, we're focusing on the proportion of specimens that have an ultimate strength of at least 34 ksi, for two different alloys, A and B.

To calculate this proportion, we divide the number of specimens meeting the criteria by the total number of specimens for each type of alloy. For alloy A, this proportion is calculated as follows: the number of specimens is 22 and the total number is 40, so the proportion is 0.55. For alloy B, with 29 specimens out of a total of 42, it comes to approximately 0.6905.

Understanding proportions is important because they give us a quantitative handle on real-world data by illustrating the part of the population that displays a particular characteristic.
Standard Error
Standard error (SE) measures the accuracy with which a sample represents a population. In the context of comparing proportions, it gives us an idea of the variability between two sample proportions.

Specifically, the standard error tells us how much the proportion estimates might vary if we were to repeat the same sample procedure. We calculate the SE for the difference in proportions using the formula: \[ SE = \sqrt{\frac{p_A(1-p_A)}{m_A} + \frac{p_B(1-p_B)}{m_B}} \] where \(p_A\) and \(p_B\) are the sample proportions, and \(m_A\) and \(m_B\) are the sample sizes for alloys A and B, respectively.

By substituting the A and B proportions and their respective populations sizes into the formula, we find an SE of approximately 0.1036. This quantifies the expected variability in the differences of proportions in samples of similar size.
Critical Value
The critical value is a key component when calculating confidence intervals. It defines the cutoff point or boundary for what we consider an unlikely event under the normal distribution.

To form a confidence interval, especially a 95% confidence interval in this case, we need to know how far out on the edges of the distribution curve our interval should extend. The critical value is usually determined based on the confidence level we want - 95% is common and corresponds to a critical value of approximately 1.96.

This 1.96 value comes from the standard normal distribution, as approximately 95% of all data units fall within 1.96 standard deviations from the mean in a bell-shaped curve. Thus, it's used to construct the confidence intervals by being multiplied with the standard error to obtain the margin of error in the proportions.
Z-score
The z-score is a statistical measurement that describes a value's position relative to the mean of a group of values, measured in terms of standard deviations. It's a normalization method to help us understand how a particular data point fits within a distribution.

For our exercise, the z-score is involved in identifying the critical value used in the confidence interval calculation. Specifically, for a 95% confidence interval, you'll often encounter the z-score of 1.96, which represents how far our interval extends from the normal distribution's mean.

Even though we don't calculate a z-score directly in this problem's steps, its concept underpins the critical value choice. By grasping z-scores, you can better appreciate how probable it is for data to occur at or beyond a certain distance from the mean under the standard normal distribution, thus helping in risk and probability assessments.

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Most popular questions from this chapter

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