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The mean pull-off force of an adhesive used in marufacturing a connector for an automotive engine application should be at least 75 pounds. This adhesive will be used unless there is strong evidence that the pull-off force does not meet this requirement. A test of an appropriate hypothesis is to be conducted with sample size \(n=10\) and \(\alpha=0.05 .\) Assume that the pull-off force is normally distributed, and \(\sigma\) is not known. (a) If the true standard deviation is \(\sigma=1\), what is the risk that the adhesive will be judged acceptable when the true mean pull-off force is only 73 pounds? Only 72 pounds? (b) What sample size is required to give a \(90 \%\) chance of detecting that the true mean is only 72 pounds when \(\sigma=1 ?\) (c) Rework parts (a) and (b) assuming that \(\sigma=2\). How much impact does increasing the value of \(\sigma\) have on the answers you obtain?

Short Answer

Expert verified
The probability of Type II error for 73 pounds is about 0.1611, and for 72 pounds, it is 0.0465. Sample size needed for 72 pounds with \( \sigma=1 \) is 12, and \( \sigma=2 \) is 45. Larger \( \sigma \) increases errors and sample size.

Step by step solution

01

Define the Null and Alternative Hypothesis

The null hypothesis is that the mean pull-off force of the adhesive, \( \mu \), satisfies \( H_0: \mu \geq 75 \) pounds. The alternative hypothesis is \( H_a: \mu < 75 \) pounds.
02

Determine the Test Statistic

Since \( \sigma \) is not known, we use the t-distribution. The test statistic \( T \) is given by \[ T = \frac{\bar{X} - \mu_0}{S / \sqrt{n}} \]where \( \bar{X} \) is the sample mean, \( S \) is the sample standard deviation, and \( \mu_0 = 75 \) pounds.
03

Calculate the Critical Value for \( n = 10 \)

With \( n = 10 \) and \( \alpha = 0.05 \), the degrees of freedom are \( n-1 = 9 \). We find the critical t-value from the t-table for \( \alpha = 0.05 \) which is approximately \( t_{0.05, 9} = -1.833 \).
04

Calculate Type II Error for \( \mu = 73 \) and \( \sigma = 1 \)

The Type II error, \( \beta \), for \( \mu = 73 \) is calculated by finding the probability that the test statistic is above \(-1.833\) when \( \mu = 73 \) pounds:\[ \beta(73) = P\left( T > -1.833 \mid \mu = 73 \right) \approx P\left( Z > \frac{-1.833 \times \frac{1}{\sqrt{10}} + 2}{1} \right) \] where \( Z \) is the standard normal variable, leading to \( \beta(73) \approx 0.1611 \).
05

Calculate Type II Error for \( \mu = 72 \) and \( \sigma = 1 \)

Similarly, determine \[ \beta(72) = P\left( Z > \frac{-1.833 \times \frac{1}{\sqrt{10}} + 3}{1} \right) \]leading to \( \beta(72) \approx 0.0465 \).
06

Calculate Sample Size for Detecting \( \mu = 72 \) and \( \sigma = 1 \)

To find \( n \) with power \( 1-\beta = 0.90 \) for \( \mu = 72 \), solve:\[ \frac{75 - 72}{\sigma / \sqrt{n}} \approx Z_{0.10} = 1.28 \] which implies \( n \approx 11.41 \). Thus, choose \( n = 12 \).
07

Evaluate \( \beta(73) \) and \( \beta(72) \) for \( \sigma = 2 \)

Repeat calculations using \( \sigma = 2 \). Calculate \[ \beta(73) = P\left( Z > \frac{-1.833 \times \frac{2}{\sqrt{10}} + 2}{2} \right) \approx 0.4131 \]\[ \beta(72) = P\left( Z > \frac{-1.833 \times \frac{2}{\sqrt{10}} + 3}{2} \right) \approx 0.2061 \].
08

Calculate \( n \) for Detecting \( \mu = 72 \) with \( \sigma = 2 \)

With \( \sigma = 2 \), find \( n \) such that:\[ \frac{75 - 72}{2/\sqrt{n}} \approx 1.28 \] gives \( n \approx 44.57 \). Choose \( n = 45 \).
09

Analyze Impact of \( \sigma \) Increase

Increasing \( \sigma \) from 1 to 2 increases Type II error probabilities and requires a larger sample size to achieve the same power. It makes detecting a true mean deviation from 75 pounds more difficult.

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Key Concepts

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

Type II Error
In hypothesis testing, Type II error (\( \beta \)) occurs when we fail to reject the null hypothesis even though it is false. It means the test results incorrectly suggest that the adhesive's pull-off force is adequate when, in fact, it isn't. Understanding and minimizing Type II error is crucial because it impacts the credibility and effectiveness of a hypothesis test. The value of Type II error depends on several factors, such as the true mean, sample size, and the standard deviation.

Consider the situation where we aim to verify if the mean pull-off force is at least 75 pounds. When the true mean falls below this threshold, say at 73 or 72 pounds, we calculate \( \beta \) for these means to gauge the risk of concluding the adhesive is acceptable when it isn't. For example, with a true mean of 73 pounds and \( \sigma = 1 \), \( \beta \) is computed as approximately 0.1611. This figure grows with increased variance indicating the greater likelihood of accepting the null hypothesis erroneously.
  • Low \( \beta \) value: More reliable results, lower risk of missing a true effect
  • High \( \beta \) value: Greater risk that the adhesive that doesn’t meet standards will be misidentified as adequate
Sample Size Calculation
The sample size (\( n \)) plays a pivotal role in hypothesis testing, directly influencing the test's power and affecting Type II error. Calculating the appropriate sample size ensures sufficient test accuracy and that the test can adequately detect deviations from the null hypothesis. The formula for sample size calculation typically involves the desired power (\( 1 - \beta \)), the significance level (\( \alpha \)), and the expected effect size.

In the context of the adhesive, if we want a 90% chance to detect when the true mean is only 72 pounds (\( \sigma = 1 \)), solving the equation \[ \frac{75 - 72}{ rac{1}{\sqrt{n}}} \approx 1.28 \\] results in approximately 12. This result implies that increasing the sample size can significantly enhance the likelihood of observing the true state under an alternative hypothesis.

The impact of increasing the variance (\( \sigma = 2 \)) elevates the required sample size to 45 for maintaining power due to the broader spread in measurement values:
  • Small sample size: Large Type II error, lower power to detect true differences
  • Appropriately large sample size: Greater confidence in test accuracy and reduced likelihood of error
T-Distribution
When conducting tests on small sample sizes or unknown population standard deviations, the t-distribution becomes the tool of choice. Unlike the normal distribution, which is used for larger sample sizes or when \( \sigma \) is known, the t-distribution accounts for increased variability.It’s typically more spread out and has heavier tails. This feature helps minimize the risk of error when drawing conclusions from smaller samples, like our scenario with \( n = 10 \).

In the given exercise, the test statistic is derived using the t-distribution:\[ T = \frac{\bar{X} - 75}{S / \sqrt{n}} \] This accounts for the sample mean (\( \bar{X} \)) versus the population mean while factoring in sample variability. The t-distribution's critical value at \( \alpha = 0.05 \) with \( n-1 = 9 \) degrees of freedom here is \( -1.833 \). Using this distribution helps ensure sound decision-making even when data size is restricted.
  • Small samples: T-distribution preferred due to more spread and heavy tails
  • Degree of freedom affects critical values, vital for hypothesis tests with limited data

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Most popular questions from this chapter

A manufacturer of semiconductor devices takes a random sample of size \(n\) of chips and tests them, classifying each chip as defective or nondefective. Let \(X_{i}=0\) if the chip is nondefective and \(X_{i}=1\) if the chip is defective. The sample fraction defective is $$ \hat{p}=\frac{X_{1}+X_{2}+\cdots+X_{n}}{n} $$ What are the sampling distribution, the sample mean, and sample variance estimates of \(\hat{p}\) when (a) The sample size is \(n=50 ?\) (b) The sample size is \(n=80 ?\) (c) The sample size is \(n=100 ?\) (d) Compare your answers to parts (a)-(c) and comment on the effect of sample size on the variance of the sampling distribution.

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