/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 Compared single versus dual spin... [FREE SOLUTION] | 91Ó°ÊÓ

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Compared single versus dual spindle saw processes for copper metallized wafers. A total of 15 devices of each type were measured for the width of the backside chipouts, \(\bar{x}_{\text {single }}=66.385, s_{\text {single }}=7.895\) and \(\bar{x}_{\text {double }}=45.278, s_{\text {double }}=8.612\). (a) Do the sample data support the claim that both processes have the same chip outputs? Use \(\alpha=0.05\) and assume that both populations are normally distributed and have the same variance. Find the \(P\) -value for the test. (b) Construct a \(95 \%\) two-sided confidence interval on the mean difference in spindle saw process. Compare this interval to the results in part (a). (c) If the \(\beta\) -error of the test when the true difference in chip outputs is 15 should not exceed \(0.1,\) what sample sizes must be used? Use \(\alpha=0.05\).

Short Answer

Expert verified
The data supports rejecting the null hypothesis with a significant difference. The confidence interval excludes 0, supporting the test result. Sample sizes need to increase to maintain a β of 0.1 if the true difference is 15.

Step by step solution

01

Understanding the Problem

We need to test if there is a significant difference between the mean chip outputs of single and dual spindle saw processes using hypothesis testing. We will conduct a two-sample t-test assuming equal variances, calculate the confidence interval for the mean difference, and determine the required sample size to limit the \(\beta\) error to not exceed 0.1.
02

Hypothesis Formulation for Part (a)

Set up the null hypothesis \( H_0: \mu_{\text{single}} = \mu_{\text{double}} \) and the alternative hypothesis \( H_1: \mu_{\text{single}} eq \mu_{\text{double}} \). This two-tailed test assumes both populations have the same variance.
03

Calculate Test Statistic for Part (a)

Use the formula for the test statistic\[ t = \frac{\bar{x}_{\text{single}} - \bar{x}_{\text{double}}}{s_p \cdot \sqrt{\frac{2}{n}}} \]where \( s_p = \sqrt{ \frac{(n - 1)s_{\text{single}}^2 + (n - 1)s_{\text{double}}^2}{2n - 2} } \) is the pooled standard deviation, and \( n = 15 \).
04

Compute the P-value for Part (a)

After calculating the test statistic \( t \), find the P-value using a t-distribution table or calculator for \( 2n - 2 = 28 \) degrees of freedom. Compare the P-value to the significance level \( \alpha = 0.05 \).
05

Decision for Hypothesis Test in Part (a)

If the P-value is less than \( \alpha = 0.05 \), reject the null hypothesis. Otherwise, do not reject it. This will tell us if the data supports the claim that the processes have the same chip outputs.
06

Constructing 95% Confidence Interval for Part (b)

Calculate the confidence interval for the difference in means using:\[ \left( (\bar{x}_{\text{single}} - \bar{x}_{\text{double}}) \pm t_{\alpha/2, 28} \cdot s_p \cdot \sqrt{\frac{2}{n}} \right) \]Where \( t_{\alpha/2, 28} \) is the critical value from the t-distribution for 28 degrees of freedom.
07

Compare Confidence Interval to Part (a) Results

Check if 0 is within the confidence interval. If 0 is outside the interval, it supports rejecting the null hypothesis, indicating a significant difference between processes.
08

Determine Sample Size for Part (c)

Use the formula for sample size determination:\[ n = \left( \frac{(z_{\alpha/2} + z_{\beta}) \cdot s_p}{\Delta} \right)^2 \]Where \( z_{\alpha/2} \) and \( z_{\beta} \) are Z-scores from the standard normal distribution for \( \alpha = 0.05 \) and \( \beta = 0.1 \), respectively, and \( \Delta = 15 \) is the true difference.

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Key Concepts

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

Two-Sample T-Test
A two-sample t-test is a statistical method used to determine if two population means are significantly different from each other. In the context of comparing single and dual spindle saw processes, this test helps us evaluate whether the difference in their output, measured through backside chipouts, indicates a true difference in the processes or if it is due to random chance.

Before conducting the test, we set up our hypotheses. The null hypothesis () states that the means are equal (<\mu_{\text{single}} = \mu_{\text{double}}>), while the alternative hypothesis () suggests they are different (<\mu_{\text{single}} eq \mu_{\text{double}}>). We use the formula for the t-test statistic: <\[ t = \frac{\bar{x}_{\text{single}} - \bar{x}_{\text{double}}}{s_p \cdot \sqrt{\frac{2}{n}}} \]>, where <\( s_p \)> is the pooled standard deviation, calculated based on the variation within each sample.

After determining the test statistic, we compare it with critical values from the t-distribution with the appropriate degrees of freedom. The P-value derived from this comparison will tell us if our test is significant at the chosen level of significance, <\alpha = 0.05>. A P-value less than <\alpha> implies rejecting the null hypothesis, suggesting the sample data does not support the claim that both processes have the same chip outputs.
Confidence Interval
The confidence interval provides a range within which the true difference in the spindle processes' means is expected to fall, with a certain level of confidence, typically 95%. It offers a more nuanced view than a binary hypothesis test result. In our exercise, constructing a 95% confidence interval helps confirm the findings of the hypothesis test by checking if this interval includes zero.

The confidence interval is calculated using the formula:<\[ \left( (\bar{x}_{\text{single}} - \bar{x}_{\text{double}}) \pm t_{\alpha/2, 28} \cdot s_p \cdot \sqrt{\frac{2}{n}} \right) \]>. Here, <\( t_{\alpha/2, 28} \)> is the critical value of the t-distribution for 28 degrees of freedom at a 95% confidence level and <\( s_p \)> is the pooled standard deviation.

If this interval does not contain zero, it supports the conclusion from hypothesis testing that there is a significant difference between the processes. It provides insight into both the direction and magnitude of the mean difference, offering more detailed information than a hypothesis test alone.
Sample Size Determination
Determining the appropriate sample size is crucial to achieving reliable statistical results. In this context, sample size needs to be calculated to control the Type II error (<\beta>), which is the probability of failing to detect a difference when one actually exists.

For the given exercise, we need the <\beta>-error to not exceed 0.1 when the true difference in means is assumed to be 15. The formula for sample size is:<\[ n = \left( \frac{(z_{\alpha/2} + z_{\beta}) \cdot s_p}{\Delta} \right)^2 \]>, where <\( z_{\alpha/2} \)> and <\( z_{\beta} \)> are the critical values from the standard normal distribution that correspond to the given <\alpha> and <\beta> levels, respectively, and <\( \Delta \)> is the anticipated true difference of 15.

This calculation ensures that the designed study is adequately powered, meaning it has a high probability of detecting the true difference when it exists, thus providing confidence in the conclusions drawn from the analysis.

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Most popular questions from this chapter

A study was performed to determine whether men and women differ in their repeatability in assembling components on printed circuit boards. Random samples of 25 men and 21 women were selected, and each subject assembled the units. The two sample standard deviations of assembly time were \(s_{\text {men }}=0.98\) minutes and \(s_{\text {women }}=1.02\) minutes. (a) Is there evidence to support the claim that men and women differ in repeatability for this assembly task? Use \(\alpha=\) 0.02 and state any necessary assumptions about the underlying distribution of the data. (b) Find a \(98 \%\) confidence interval on the ratio of the two variances. Provide an interpretation of the interval.

Two different formulations of an oxygenated motor fuel are being tested to study their road octane numbers. The variance of road octane number for formulation 1 is \(\sigma_{1}^{2}=\) \(1.5,\) and for formulation 2 it is \(\sigma_{2}^{2}=1.2 .\) Two random samples of size \(n_{1}=15\) and \(n_{2}=20\) are tested, and the mean road octane numbers observed are \(\bar{x}_{1}=89.6\) and \(\bar{x}_{2}=92.5\). Assume normality. (a) If formulation 2 produces a higher road octane number than formulation \(1,\) the manufacturer would like to detect it. Formulate and test an appropriate hypothesis, using \(\alpha=0.05 .\) What is the \(P\) -value? (b) Explain how the question in part (a) could be answered with a \(95 \%\) confidence interval on the difference in mean road octane number. (c) What sample size would be required in each population if we wanted to be \(95 \%\) confident that the error in estimating the difference in mean road octane number is less than \(1 ?\)

Two machines are used to fill plastic bottles with dishwashing detergent. The standard deviations of fill volume are known to be \(\sigma_{1}=0.10\) fluid ounces and \(\sigma_{2}=0.15\) fluid ounces for the two machines, respectively. Two random samples of \(n_{1}=12\) bottles from machine 1 and \(n_{2}=10\) bottles from machine 2 are selected, and the sample mean fill volumes are \(\bar{x}_{1}=30.87\) fluid ounces and \(\bar{x}_{2}=30.68\) fluid ounces. Assume normality. (a) Construct a \(90 \%\) two-sided confidence interval on the mean difference in fill volume. Interpret this interval. (b) Construct a \(95 \%\) two-sided confidence interval on the mean difference in fill volume. Compare and comment on the width of this interval to the width of the interval in part (a). (c) Construct a \(95 \%\) upper-confidence interval on the mean difference in fill volume. Interpret this interval. (d) Test the hypothesis that both machines fill to the same mean volume. Use \(\alpha=0.05 .\) What is the \(P\) -value? (e) If the \(\beta\) -error of the test when the true difference in fill volume is 0.2 fluid ounces should not exceed \(0.1,\) what sample sizes must be used? Use \(\alpha=0.05 .\)

Suppose that we wish to test \(H_{0}: \mu=\mu_{0}\) versus \(H_{1}: \mu \neq \mu_{0},\) where the population is normal with known \(\sigma\). Let \(0<\epsilon<\alpha,\) and define the critical region so that we will reject \(H_{0}\) if \(z_{0}>z_{\epsilon}\) or if \(z_{0}<-z_{\alpha-\epsilon},\) where \(z_{0}\) is the value of the usual test statistic for these hypotheses. (a) Show that the probability of type I error for this test is \(\alpha\). (b) Suppose that the true mean is \(\mu_{1}=\mu_{0}+\Delta\). Derive an expression for \(\beta\) for the above test.

In a random sample of 200 Phoenix residents who drive a domestic car, 165 reported wearing their seat belt regularly, while another sample of 250 Phoenix residents who drive a foreign car revealed 198 who regularly wore their seat belt. (a) Perform a hypothesis-testing procedure to determine if there is a statistically significant difference in seat belt usage between domestic and foreign car drivers. Set your probability of a type I error to \(0.05 .\) (b) Perform a hypothesis-testing procedure to determine if there is a statistically significant difference in seat belt usage between domestic and foreign car drivers. Set your probability of a type I error to 0.1 (c) Compare your answers for parts (a) and (b) and explain why they are the same or different. (d) Suppose that all the numbers in the problem description were doubled. That is, in a random sample of 400 Phoenix residents who drive a domestic car, 330 reported wearing their seat belt regularly, while another sample of 500 Phoenix residents who drive a foreign car revealed 396 who regularly wore their seat belt. Repeat parts (a) and (b) and comment on the effect of increasing the sample size without changing the proportions on your results.

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