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Suppose the expected tensile strength of type-A steel is \(105 \mathrm{ksi}\) and the standard deviation of tensile strength is \(8 \mathrm{ksi}\). For type-B steel, suppose the expected tensile strength and standard deviation of tensile strength are \(100 \mathrm{ksi}\) and \(6 \mathrm{ksi}\), respectively. Let \(\bar{X}=\) the sample average tensile strength of a random sample of 40 type-A specimens, and let \(\bar{Y}=\) the sample average tensile strength of a random sample of 35 type-B specimens. a. What is the approximate distribution of \(\bar{X}\) ? Of \(\bar{Y}\) ? b. What is the approximate distribution of \(\bar{X}-\bar{Y}\) ? Justify your answer. c. Calculate (approximately) \(P(-1 \leq \bar{X}-\bar{Y} \leq 1)\). d. Calculate \(P(\bar{X}-\bar{Y} \geq 10)\). If you actually observed \(\bar{X}-\bar{Y} \geq 10\), would you doubt that \(\mu_{1}-\mu_{2}=5\) ?

Short Answer

Expert verified
\(\bar{X} \sim N(105, 1.26); \bar{Y} \sim N(100, 1.01); \bar{X}-\bar{Y} \sim N(5, 1.62)\). Probability of \(-1 \leq \bar{X}-\bar{Y} \leq 1\) is approximately \(0.01\). If \(\bar{X}-\bar{Y} \geq 10\), doubt that \(\mu_1 - \mu_2 = 5\).

Step by step solution

01

Approximate Distribution of \(\bar{X}\)

Given the expected tensile strength \(\mu = 105\) ksi and the standard deviation \(\sigma = 8\) ksi for type-A steel, and sample size \(n = 40\), the distribution of \(\bar{X}\) can be approximated using the Central Limit Theorem (CLT). Thus, \(\bar{X} \sim N(105, \frac{8}{\sqrt{40}})\), where \(\frac{8}{\sqrt{40}}\) is the standard error.
02

Approximate Distribution of \(\bar{Y}\)

For type-B steel, with \(\mu = 100\) ksi and \(\sigma = 6\) ksi, and sample size \(n = 35\), the distribution of \(\bar{Y}\) is \(\bar{Y} \sim N(100, \frac{6}{\sqrt{35}})\) because of the CLT. Here, \(\frac{6}{\sqrt{35}}\) is the standard error.
03

Approximate Distribution of \(\bar{X}-\bar{Y}\)

The difference \(\bar{X} - \bar{Y}\) also follows a normal distribution because both \(\bar{X}\) and \(\bar{Y}\) are normally distributed. The mean is the difference of their means: \(105 - 100 = 5\). The variance is the sum of their variances: \(\left(\frac{8}{\sqrt{40}}\right)^2 + \left(\frac{6}{\sqrt{35}}\right)^2\). Thus, \(\bar{X} - \bar{Y} \sim N(5, \sqrt{\left(\frac{8}{\sqrt{40}}\right)^2 + \left(\frac{6}{\sqrt{35}}\right)^2})\).
04

Calculate \(P(-1 \leq \bar{X}-\bar{Y} \leq 1)\)

Find the probability for \(\bar{X} - \bar{Y}\) between -1 and 1. Standardize variable: \(Z = \frac{\bar{X} - \bar{Y} - 5}{\sqrt{\left(\frac{8}{\sqrt{40}}\right)^2 + \left(\frac{6}{\sqrt{35}}\right)^2}}\). Calculate \(P(Z \leq \frac{1-5}{\text{SE}}) - P(Z \leq \frac{-1-5}{\text{SE}})\) using Z-tables.
05

Calculate \(P(\bar{X}-\bar{Y} \geq 10)\)

Convert to \(Z\) score: \(Z = \frac{10 - 5}{\text{SE}}\). Find \(P(Z \geq \text{computed } Z)\) using the Z-table. If \(P\) is small, observing \(\bar{X}-\bar{Y} \geq 10\) indicates doubting \(\mu_1 - \mu_2 = 5\). If \(P\) is close to 1, it is consistent.

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Key Concepts

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

Normal Distribution
The concept of a normal distribution is essential in understanding the behavior of many natural phenomena, including tensile strength. At its core, a normal distribution is a type of continuous probability distribution that is symmetrical around its mean. It is also known as the Gaussian distribution. This bell-shaped curve is defined by two parameters: the mean and the standard deviation.
  • The mean dictates where the center of the curve is located. For instance, a mean tensile strength of 105 ksi indicates that 105 ksi is the most probable tensile strength for type-A steel.
  • The standard deviation informs us about the spread or dispersion of the data around the mean. A smaller standard deviation results in a steeper curve, indicating that data points are closer to the mean.
Due to the Central Limit Theorem, we know that the distribution of sample means will approximate a normal distribution as the sample size increases, even if the original variables themselves are not normally distributed. This allows us to easily predict the distribution of tensile strength averages for samples of steels.
Tensile Strength
Tensile strength is essentially the maximum amount of tensile (stretching) force that a material can withstand before failure. In the context of the steel types mentioned, tensile strength is a critical property because it defines how strong the steel is before it yields or breaks. For type-A steel, an expected tensile strength of 105 ksi implies that, on average, this is the maximum stress it can endure per square inch. Similarly, type-B steel, with a 100 ksi expected tensile strength, is slightly less robust. These values help engineers and designers in selecting the appropriate type of steel for various applications, ensuring structural soundness and integrity. When testing steel, samples are often taken, and their tensile strengths are measured. These sample data allow for comparing steel types or evaluating quality consistency, thereby playing a fundamental role in manufacturing and construction.
Standard Deviation
Standard deviation is a statistical measure used to quantify the amount of variation or dispersion in a set of data values. In the context of tensile strength, it tells us how much the tensile strengths of different samples deviate from the average tensile strength.
  • For type-A steel, a standard deviation of 8 ksi indicates that the tensile strengths generally vary 8 ksi above or below its mean of 105 ksi.
  • For type-B steel, a smaller standard deviation of 6 ksi means its tensile strength numbers are more closely clustered around its mean of 100 ksi.
The standard deviation is crucial because it helps determine the material's reliability. A smaller standard deviation suggests that a material will perform more consistently, which is often desired in engineering fields, as it assures predictable performance under stress.
Sample Size
Sample size refers to the number of observations or data points collected in a study or experiment. It plays a significant role in the accuracy and reliability of the statistical results. In the case of the tensile strength study for type-A and type-B steels, sample sizes of 40 and 35 are used respectively.
  • A larger sample size generally provides greater confidence in the computed mean and more accurately reflects the population mean. This is because larger samples reduce the effect of outliers and random variation.
  • With a sample size of 40 for type-A steel and 35 for type-B steel, the Central Limit Theorem can apply, allowing the sample means (\( \bar{X} \) for type-A and \( \bar{Y} \) for type-B) to be approximated as normal distributions.
The accuracy of predictions regarding tensile strength distributions increases with larger sample sizes, providing a more stable basis for comparing the two types of steel and drawing informed conclusions about their characteristics and performance.

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