91Ó°ÊÓ

An engineer who is studying the tensile strength of a steel alloy intended for use in golf club shafts knows that tensile strength is approximately normally distributed with \(\sigma=60\) psi. A random sample of 12 specimens has a mean tensile strength of \(\bar{x}=3450\) psi. (a) Test the hypothesis that mean strength is 3500 psi. Use \(\alpha=0.01\) (b) What is the smallest level of significance at which you would be willing to reject the null hypothesis? (c) What is the \(\beta\) -error for the test in part (a) if the true mean is \(3470 ?\) (d) Suppose that you wanted to reject the null hypothesis with probability at least 0.8 if mean strength \(\mu=3470 .\) What sample size should be used? (e) Explain how you could answer the question in part (a) with a two-sided confidence interval on mean tensile strength.

Step by step solution

01

Define Hypotheses for Part (a)

State the null hypothesis \( H_0 \) and the alternative hypothesis \( H_a \)\[ H_0: \mu = 3500 \]\[ H_a: \mu eq 3500 \]

02

Determine the Test Statistic for Part (a)

The formula for the test statistic \( z \) is:\[ z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} \]Substitute the given values:\[ \bar{x} = 3450, \ \mu = 3500, \ \sigma = 60, \ n = 12 \]Calculate:\[ z = \frac{3450 - 3500}{60/\sqrt{12}} = \frac{-50}{60/3.464} \approx -2.886 \]

03

Determine the Critical Value for Part (a)

For a two-tailed test with \( \alpha = 0.01 \), the critical z-values are \( \pm z_{\alpha/2} = \pm z_{0.005} = \pm 2.576 \).

04

Decision Rule for Part (a)

Since \( z = -2.886 \)< \( -2.576 \), we reject the null hypothesis.

05

Calculate p-Value for Part (b)

Calculate the p-value: \( 2 \times P(Z < -2.886) \). Find \( P(Z < -2.886) \) using a standard normal distribution table or calculator:- \( P(Z < -2.886) \approx 0.0019 \).- Therefore, \( \text{p-value} = 2 \times 0.0019 = 0.0038 \).The smallest significance level at which we reject \( H_0 \) is \( \alpha = 0.0038 \).

06

Determine \( \beta \)-Error for Part (c)

\( \beta \) is the probability of failing to reject \( H_0 \) when \( \mu = 3470 \).Calculate new \( z \) using \( \mu = 3470 \):\[ z = \frac{3470 - 3500}{60/\sqrt{12}} = \frac{-30}{17.32} \approx -1.732 \]Find \( P(Z > \text{upper critical value}) - P(Z < \text{lower critical value}) \):\( P(Z > 2.576) - P(Z < -1.732) \).\( \beta \approx P(Z > 2.576) + P(Z < -1.732) \).Use \( P(Z > 2.576) \approx 0.005 \) and \( P(Z < -1.732) \approx 0.0427 \).Thus, \( \beta \approx 0.005 + 0.0427 = 0.0477 \).

07

Calculate Sample Size for Part (d)

If \( \mu = 3470 \) and the test needs to be significant, \( Z_{\beta} = 0.8416 \) for power 0.8 and \( Z_{\alpha/2} = 2.576 \).Use the formula:\[ n = \left( \frac{Z_{\alpha/2} \cdot \sigma + Z_{\beta} \cdot \sigma}{\mu_0 - \mu_a} \right)^2 \]\[ n = \left( \frac{2.576 \cdot 60 + 0.8416 \cdot 60}{3500 - 3470} \right)^2 \]Calculate:\[ n = \left( \frac{154.56 + 50.496}{30} \right)^2 \approx 43.26 \]Thus, \( n \) must be at least 44.

08

Confidence Interval for Part (e)

Calculate a 99% confidence interval for \( \mu \):\[ \bar{x} \pm z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}} \]\[ 3450 \pm 2.576 \times \frac{60}{3.464} \]\[ 3450 \pm 44.64 \]The interval is \([3405.36, 3494.64]\).Since 3500 is not in this interval, we reject \( H_0 \).

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Key Concepts

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

Tensile Strength

Tensile strength is a critical property of materials, especially metals like steel, determining how much pulling (tensile) force a material can withstand before failure. This property is particularly essential for applications where the material will undergo significant stress, such as in the construction of golf club shafts.
Tensile strength can be understood through testing, where a sample material is subjected to a controlled tension until it breaks. The maximum stress experienced by the material is then recorded.
Understanding tensile strength is crucial for ensuring that materials are fit for their intended purpose, preventing structural failures which can be costly or dangerous.

Normal Distribution

In the context of the problem, normal distribution is a statistical concept that describes how data points are distributed around a mean. It is often called the 'bell curve' due to its shape.
When we say that tensile strength is normally distributed, it means that most of the tensile strength measurements tend to cluster around the average (mean) value, with fewer measurements appearing as you move away from the mean in either direction.

• The normal distribution is characterized by two parameters: the mean (\(\mu\)) and the standard deviation (\(\sigma\)).
• The mean is the average value, and the standard deviation measures how spread out the values are around the mean.

In hypothesis testing, the assumption of normal distribution allows us to use certain statistical tests, like the z-test, which makes it easier to make inferences about the population based on sample data.

Confidence Interval

A confidence interval is a range of values, derived from the sample data, that is likely to contain the true population parameter, such as the mean tensile strength.
It gives us an estimated range for the mean rather than a single number, providing a degree of certainty about our estimation. In simple terms, it's a way to quantify the uncertainty of an estimate.

• A 99% confidence interval means that if we were to take repeated samples and build an interval each time, 99% of those intervals would contain the true mean tensile strength.
• The width of the interval depends on the sample size and variability in the data.

For instance, in the problem provided, the confidence interval calculated was [3405.36, 3494.64]. This interval gives insight into where the true mean tensile strength could lie, and since it does not include the hypothesized mean of 3500 psi, it supports rejecting the null hypothesis. With the help of confidence intervals, we can support decision-making in engineering and quality control.