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Chapter 10: Problem 55

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.

Short Answer

Expert verified
No evidence supports differing repeatability; variances are likely equal.

Step by step solution

01

State the Hypotheses for Part (a)

We need to determine if there is evidence for a difference in repeatability between men and women. This involves a two-tailed F-test on variances. The null hypothesis is that the variances are equal: \(H_0: \sigma^2_{\text{men}} = \sigma^2_{\text{women}}\). The alternative hypothesis is that they are not equal: \(H_a: \sigma^2_{\text{men}} eq \sigma^2_{\text{women}}\).
02

Define the Test Statistic for Part (a)

The test statistic for comparing two variances is the F-statistic, which is given by: \( F = \frac{s^2_{\text{women}}}{s^2_{\text{men}}} \) if \( s^2_{\text{women}} > s^2_{\text{men}} \), otherwise flip the ratio. This is because we always divide the larger variance by the smaller one to get an F-distribution with appropriate degrees of freedom.
03

Calculate the Test Statistic for Part (a)

Calculate the F-statistic using the sample standard deviations: \( F = \frac{1.02^2}{0.98^2} \approx 1.0825 \).
04

Determine the Critical F-value for Part (a)

The degrees of freedom for men and women are \( df_{\text{men}} = 24 \) and \( df_{\text{women}} = 20 \). For a two-tailed test with \( \alpha = 0.02 \), find the critical F-values using an F-distribution table or calculator: \( F_{\text{critical}} = (F_{0.01}, F_{0.99}) = (2.54, 0.392) \).
05

Decision for Part (a)

The calculated F-statistic (1.0825) is not in the critical region beyond the limits 0.392 or 2.54. Therefore, we do not reject the null hypothesis, indicating no significant difference in variance between men and women at \( \alpha = 0.02 \) level.
06

State Assumptions for Part (a)

Assume the data is normally distributed and the samples are independent to apply the F-test for two variances.
07

Confidence Interval for Part (b) - Calculate the Interval

For a confidence interval on the ratio of variances, use the formula: \( (\frac{s^2_{\text{women}}}{s^2_{\text{men}}} \cdot \frac{1}{F'_{\alpha/2}}, \frac{s^2_{\text{women}}}{s^2_{\text{men}}} \cdot F'_{\alpha/2}) \). Calculate using the F-distribution values from before: \( CI = (1.0825 \cdot \frac{1}{2.54}, 1.0825 \cdot 2.54) = (0.426, 2.75) \).
08

Interpret the Confidence Interval for Part (b)

The 98% confidence interval for the ratio of variances is (0.426, 2.75). This interval includes 1, suggesting there is no significant evidence to suggest the variances and thus the repeatability are different between men and women.

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Key Concepts

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

F-test
The F-test is a statistical method used to compare two variances to test if they are significantly different from each other. It is commonly applied when you want to compare the variances of two independent samples, like in this study where we are looking at assembly times of men versus women. An F-test is performed by calculating the F-statistic, which is the ratio of the two sample variances. In the formula:
  • F = \( \frac{s^2_{\text{women}}}{s^2_{\text{men}}} \)
We use the larger variance in the numerator to ensure that the ratio is always greater than or equal to 1. This enables the use of the F-distribution, which is a skewed distribution used in hypothesis testing.
The critical value for the F-test is derived from an F-distribution table, based on the degrees of freedom of the samples and the significance level \(\alpha\). If the calculated F-statistic lies outside the range of critical values, the null hypothesis (that variances are equal) can be rejected. In this exercise, the F-statistic was found not to exceed the critical boundaries, suggesting that the variances in assembly times for men and women are not significantly different. All statistical tests, including the F-test, require assumptions to be checked before interpretation.
Variance Analysis
Variance analysis is crucial in understanding the dispersion of different datasets. In hypothesis testing, it is used to assess whether the variability of two populations are statistically different. For instance, comparing the variances in assembly times highlights the consistency in performance between men and women in the study.
  • The calculation involves the sample standard deviations converted to variance by squaring them.
  • A hypothesis test like the F-test might then be used to compare these variances.
In essence, variance analysis helps to identify differences in consistency or spread within datasets. A small variance implies data points are close to the mean and thus similar, indicating high repeatability. Conversely, a large variance indicates more spread out data. This exercise suggests no major difference in variances between our two groups.
Confidence Interval
A confidence interval is a range estimate around a sample statistic that provides an indication of how much variability there is likely to be in that statistic across different samples. In this exercise, you calculated a confidence interval for the ratio of the two population variances - men and women's assembly times. With a 98% confidence level, it suggests how certain we are that a true ratio lies within this range.
  • The confidence interval was found using the formula: \( (\frac{s^2_{\text{women}}}{s^2_{\text{men}}} \cdot \frac{1}{F'_{\alpha/2}}, \frac{s^2_{\text{women}}}{s^2_{\text{men}}} \cdot F'_{\alpha/2}) \), incorporating F-distribution values.
  • The resulting interval from the solution, (0.426, 2.75), includes 1, which infers there is no definitive evidence of different variances as this value indicates no preference towards greater variance in either group.
Confidence intervals offer a measure of precision and risk assessment for statistical outcomes.
Statistical Assumptions
Successful implementation of hypothesis tests and confidence intervals depends heavily on certain statistical assumptions. These assumptions must be verified to ensure that the results from statistical analyses are valid and reliable. Key assumptions include:
  • Normal distribution: The data from each group needs to be normally distributed, which means that the data follows a symmetric, bell-shaped distribution.
  • Independence: The samples (in this case, men and women) must be independent of each other, implying that the data points do not influence one another.
  • Equal variance: For some tests, we assume variances are equal across groups, though in this case, our goal is testing them for difference.
Failing to meet these assumptions can lead to incorrect conclusions. Therefore, before proceeding with analyses such as the F-test, it is crucial to check these assumptions. In this study, the assumption of normal distribution and independence was made for applying the F-test. These include the backbone of statistical testing, ensuring rigorous and valid conclusions.

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Most popular questions from this chapter

Two machines are used for filling plastic bottles with a net volume of 16.0 ounces. The fill volume can be assumed normal, with standard deviation \(\sigma_{1}=0.020\) and \(\sigma_{2}=0.025\) ounces. A member of the quality engineering staff suspects that both machines fill to the same mean net volume, whether or not this volume is 16.0 ounces. A random sample of 10 bottles is taken from the output of each machine. (a) Do you think the engineer is correct? Use \(\alpha=0.05 .\) What is the \(P\) -value for this test? (b) Calculate a \(95 \%\) confidence interval on the difference in means. Provide a practical interpretation of this interval. (c) What is the power of the test in part (a) for a true difference in means of \(0.04 ?\) (d) Assuming equal sample sizes, what sample size should be used to assure that \(\beta=0.05\) if the true difference in means is 0.04 ? Assume that \(\alpha=0.05\).

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 \(\alpha=0.05\).

Considered arthroscopic meniscal repair with an absorbable screw. Results showed that for tears greater than 25 millimeters, 14 of 18 \((78 \%)\) repairs were successful while for shorter tears, 22 of 30 (73\%) repairs were successful. (a) Is there evidence that the success rate is greater for longer tears? Use \(\alpha=0.05 .\) What is the \(P\) -value? (b) Calculate a one-sided \(95 \%\) confidence bound on the difference in proportions that can be used to answer the question in part (a).

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 \(\alpha=0.05 .\)
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