/*! 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 69 A study was performed to determi... [FREE SOLUTION] | 91Ó°ÊÓ

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A study was performed to determine whether men and women differ in 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 \(s_{\text {wamen }}=1.02\) 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
There is no evidence that men and women differ in repeatability. The 98% confidence interval for the variance ratio is (0.364, 2.348), supporting no difference in variance.

Step by step solution

01

State the Hypotheses

We need to test if there is a difference in the variances of assembly times between men and women. The null hypothesis is that the variances are equal, \[ H_0: \sigma^2_{\text{men}} = \sigma^2_{\text{women}} \]. The alternative hypothesis states that the variances are not equal, \[ H_1: \sigma^2_{\text{men}} eq \sigma^2_{\text{women}} \].
02

Choose the Test and Critical Value

We'll use the F-test to compare the variances because it's designed for two-sample variance comparison. Let the significance level be \( \alpha = 0.02 \). We will compute the critical F-value using degrees of freedom \( (24, 20) \) with \( \alpha/2 = 0.01 \) at each tail.
03

Calculate the F-statistic

The formula for the F-statistic comparing variances is \[ F = \frac{s^2_{\text{men}}}{s^2_{\text{women}}} \]. Therefore, \[ F = \frac{0.98^2}{1.02^2} \approx 0.9245 \].
04

Decision Rule and Conclusion (Hypothesis Test)

Look up the critical F-values for \( (24, 20) \) at a significance level of 0.01 for both tails. If \( F \) falls outside these values, reject \( H_0 \). If not, fail to reject \( H_0 \). Assuming critical F-values are approximately \( F_{0.01, 24, 20} = 2.54 \) and \( F_{0.99, 24, 20} = 0.393 \), since \( 0.9245 \) is within the range, we fail to reject \( H_0 \). There is no evidence to suggest a difference.
05

Construct the Confidence Interval for Variance Ratio

The \( 100(1 - \alpha)\% \) confidence interval for the variance ratio is given by \[ \left(\frac{s^2_{\text{men}}}{s^2_{\text{women}}} \right) \times \frac{1}{F_{\alpha/2, 24, 20}}, \left(\frac{s^2_{\text{men}}}{s^2_{\text{women}}} \right) \times F_{1-\alpha/2, 24, 20} \]. Calculate using critical F-values, such as 0.393 and 2.54, and find the interval \[ \left(0.9245 / 2.54, 0.9245 \times 2.54\right) \approx (0.364, 2.348) \].
06

Interpret the Confidence Interval

The interval \((0.364, 2.348)\) suggests that the ratio of the variances could reasonably be any value within this range at a 98% confidence level. Since this interval includes 1, it supports the hypothesis that the variances are equal.

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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 population variances. It is particularly useful when you're dealing with two samples and want to determine if their variances are significantly different. In our scenario, we are comparing the variability in assembly times between men and women. By calculating the F-statistic, we evaluate the ratio of two sample variances against a set threshold, determined by a critical F-value. The formula for the F-statistic is given by:
  • \( F = \frac{s^2_{\text{men}}}{s^2_{\text{women}}} \)
Here, the square values (\( s^2 \)) represent the variances of the sample groups. The decision to accept or reject the null hypothesis (\( H_0 \)) that the variances are equal is based on whether our calculated F-statistic falls within a range set by these critical F-values. Understanding and applying the F-test allows us to make meaningful inferences about population variances.
confidence interval
A confidence interval provides a range of values that is likely to contain a population parameter with a certain level of confidence. In the context of our problem, we are interested in a 98% confidence interval for the ratio of the variances of assembly times. To construct this interval, we utilize the calculated F-statistic and the critical F-values for our chosen significance level.
  • The formula used to compute the confidence interval for the variance ratio is:\[ \left(\frac{s^2_{\text{men}}}{s^2_{\text{women}}} \right) \times \frac{1}{F_{\alpha/2, 24, 20}}, \left(\frac{s^2_{\text{men}}}{s^2_{\text{women}}} \right) \times F_{1-\alpha/2, 24, 20} \]
The resulting interval, such as \((0.364, 2.348)\), implies that at a 98% confidence level, the true variance ratio could be any value between these two numbers. If 1 is within this range, as it is in our case, it indicates that the sample variances may not be significantly different.
variances comparison
Comparing variances is a crucial step in determining if two sample groups exhibit similar levels of variability. This is particularly relevant in fields where consistency is critical, such as in manufacturing or quality control. The variances of a dataset describe the spread or dispersion of the data points, and when comparing these across groups, we employ statistical tests like the F-test. In our exercise, the comparison aims to identify whether the repeatability of men and women in assembling components is similar, by examining the spread of their assembly times. If the F-statistic indicates that one group's variance significantly differs from the other, we would infer that they are not equally consistent. However, if the calculated confidence interval contains 1, this suggests that the differences are not statistically significant, as seen in our given exercise.
significance level
The significance level, denoted by \( \alpha \), is a threshold used to determine the probability of rejecting the null hypothesis when it is true. It is a critical concept in hypothesis testing as it defines the risk level we are willing to take for a type I error (false positive). In this exercise, the significance level is set at 0.02, meaning there is a 2% risk of concluding that the variances are different when they are actually not.A lower significance level indicates a stricter criterion for evidence against the null hypothesis, often resulting in more reliable results. With our significance level of 0.02, we compute the critical F-values for two tails of the distribution. The final decision to reject or not reject \( H_0 \) is based on whether our F-statistic exceeds these critical values, providing a clear measure for statistical decision-making.

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Most popular questions from this chapter

An electrical engineer must design a circuit to deliver the maximum amount of current to a display tube to achieve sufficient image brightness. Within her allowable design constraints, she has developed two candidate circuits and tests prototypes of each. The resulting data (in microamperes) are as follows: $$\begin{array}{l|l}\text { Circuit 1: } & 251,255,258,257,250,251,254,250,248 \\\\\hline \text { Circuit 2: } & 250,253,249,256,259,252,260,251\end{array}$$

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The concentration of active ingredient in a liquid laundry detergent is thought to be affected by the type of catalyst used in the process. The standard deviation of active concentration is known to be 3 grams per liter regardless of the catalyst type. Ten observations on concentration are taken with each catalyst, and the data follow: $$\begin{array}{l}\text { Catalyst } 1: 57.9,66.2,65.4,65.4,65.2,62.6,67.6,63.7, \\\\\quad 67.2,71.0\end{array}$$ Catalyst 2: 66.4,71.7,70.3,69.3,64.8,69.6,68.6,69.4 $$65.3,68.8$$ (a) Find a \(95 \%\) confidence interval on the difference in mean active concentrations for the two catalysts. Find the \(P\) -value. (b) Is there any evidence to indicate that the mean active concentrations depend on the choice of catalyst? Base your answer on the results of part (a). (c) Suppose that the true mean difference in active concentration is 5 grams per liter. What is the power of the test to detect this difference if \(\alpha=0.05 ?\) (d) If this difference of 5 grams per liter is really important, do you consider the sample sizes used by the experimenter to be adequate? Does the assumption of normality seem reasonable for both samples?

The "spring-like effect" in a golf club could be determined by measuring the coefficient of restitution (the ratio of the outbound velocity to the inbound velocity of a golf ball fired at the clubhead). Twelve randomly selected drivers produced by two clubmakers are tested and the coefficient of restitution measured. The data follow: Club 1: 0.8406,0.8104,0.8234,0.8198,0.8235,0.8562 $$0.8123,0.7976,0.8184,0.8265,0.7773,0.7871$$ Club 2: 0.8305,0.7905,0.8352,0.8380,0.8145,0.8465 0.8244,0.8014,0.8309,0.8405,0.8256,0.8476 (a) Is there evidence that coefficient of restitution is approximately normally distributed? Is an assumption of equal variances justified? (b) Test the hypothesis that both brands of clubs have equal mean coefficient of restitution. Use \(\alpha=0.05 .\) What is the P-value of the test? (c) Construct a \(95 \%\) two-sided \(\mathrm{Cl}\) on the mean difference in coefficient of restitution for the two brands of golf clubs. (d) What is the power of the statistical test in part (b) to detect a true difference in mean coefficient of restitution of \(0.2 ?\) (e) What sample size would be required to detect a true difference in mean coefficient of restitution of 0.1 with power of approximately \(0.8 ?\)

Consider the hypothesis test \(H_{0}: \mu_{1}=\mu_{2}\) against \(H_{1}: \mu_{1} \neq \mu_{2} .\) Suppose that sample sizes are \(n_{1}=15\) and \(n_{2}=15\) that \(\bar{x}_{1}=4.7\) and \(\bar{x}_{2}=7.8,\) and that \(s_{1}^{2}=4\) and \(s_{2}^{2}=6.25 .\) Assume that \(\sigma_{1}^{2}=\sigma_{2}^{2}\) and that the data are drawn from normal distributions. Use \(\alpha=0.05\). (a) Test the hypothesis and find the \(P\) -value. (b) Explain how the test could be conducted with a confidence interval. (c) What is the power of the test in part (a) for a true difference in means of \(3 ?\) (d) Assume that sample sizes are equal. What sample size should be used to obtain \(\beta=0.05\) if the true difference in means is -2 ? Assume that \(\alpha=0.05\).

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