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Two suppliers manufacture a plastic gear used in a laser printer. The impact strength of these gears measured in foot-pounds is an important characteristic. A random sample of 10 gears from supplier 1 results in \(\bar{x}_{1}=290\) and \(s_{1}=12\), and another random sample of 16 gears from the second supplier results in \(\bar{x}_{2}=321\) and \(s_{2}=22\) a. Is there evidence to support the claim that supplier 2 provides gears with higher mean impact strength? Use \(\alpha=0.05,\) and assume that both populations are normally distributed but the variances are not equal. What is the \(P\) -value for this test? b. Do the data support the claim that the mean impact strength of gears from supplier 2 is at least 25 foot-pounds higher than that of supplier \(1 ?\) Make the same assumptions as in part (a). c. Construct a confidence interval estimate for the difference in mean impact strength, and explain how this interval could be used to answer the question posed regarding supplier-to-supplier differences.

Short Answer

Expert verified
Part a: Reject \( H_0 \), Supplier 2 has higher strength; \( p \approx 0.0001 \). Part b: No evidence mean difference is at least 25. Part c: Difference is between 17.03 and 44.97.

Step by step solution

01

Setting Up the Hypothesis for Part a

We need to test the null hypothesis \( H_0: \mu_1 = \mu_2 \) against the alternative hypothesis \( H_a: \mu_1 < \mu_2 \), where \( \mu_1 \) and \( \mu_2 \) are the mean impact strengths for suppliers 1 and 2, respectively. Since we are assuming that the variances are not equal, we will use a two-sample t-test with unequal variances (Welch's t-test).
02

Calculating the Test Statistic for Part a

The test statistic is calculated using the formula:\[t = \frac{\bar{x}_2 - \bar{x}_1}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} = \frac{321 - 290}{\sqrt{\frac{12^2}{10} + \frac{22^2}{16}}}\]Compute the terms step-by-step, \[ = \frac{31}{\sqrt{14.4 + 30.25}} = \frac{31}{6.666} \approx 4.65\]
03

Determining the Degrees of Freedom

The degrees of freedom for Welch's t-test is calculated using the formula:\[ df = \frac{ \left( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right)^2 }{ \frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1} }\]Computing each term,\[ df \approx \frac{(14.4 + 30.25)^2}{\frac{(14.4)^2}{9} + \frac{(30.25)^2}{15}} \approx 19.225 \approx 19\]
04

Finding the Critical Value and Decision for Part a

Using a t-distribution table or calculator, find the critical t-value for a one-tailed test with \( \alpha = 0.05 \) at \( df = 19 \). The critical t-value is approximately 1.729. Since \( 4.65 > 1.729 \), we reject the null hypothesis.
05

Calculating the p-value for Part a

Using a calculator or software to find the p-value for \( t = 4.65 \) with \( df = 19 \), we find that \( p \approx 0.0001 \), which supports rejecting the null hypothesis. Supplier 2 seems to have a higher mean impact strength.
06

Setting Up the Hypothesis for Part b

For part b, test the null hypothesis \( H_0: \mu_2 - \mu_1 = 25 \) against the alternative hypothesis \( H_a: \mu_2 - \mu_1 > 25 \). Use the same t-test setup as before.
07

Calculating the Test Statistic for Part b

The test statistic for part b is:\[t = \frac{(\bar{x}_2 - \bar{x}_1) - 25}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} = \frac{321 - 290 - 25}{6.666} = \frac{6}{6.666} \approx 0.9\]
08

Decision for Part b

The critical t-value is still about 1.729. Since \( 0.9 < 1.729 \), we fail to reject the null hypothesis. The data does not support the claim that the mean is at least 25 foot-pounds higher.
09

Constructing the Confidence Interval for Part c

Construct a confidence interval for \( \mu_2 - \mu_1 \) using the formula:\[CI = (\bar{x}_2 - \bar{x}_1) \pm t_{df, \alpha/2} \times \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}\]Using \( 95\% \) confidence and \( df = 19 \), find the value of \( t \approx 2.093 \):\[CI = 31 \pm 2.093 \times 6.666 \Rightarrow CI = (17.03, 44.97)\]
10

Interpretation of the Confidence Interval

The confidence interval for the difference in mean impact strengths is \((17.03, 44.97)\). Since 25 is within this interval, we cannot confidently say the difference is at least 25, aligning with our earlier result in part b.

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Key Concepts

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

Welch's t-Test
Welch's t-Test is a type of statistical test used to determine if there is a significant difference between the means of two independent samples. Unlike the traditional two-sample t-test, Welch's t-test does not assume equal variances between the samples. This makes it a more robust choice, especially when dealing with datasets that have unequal variances.

In the exercise, we apply Welch's t-test to evaluate the impact strength of plastic gears from two different suppliers. We formulate the null hypothesis (\( H_0: \mu_1 = \mu_2 \) ) that suggests there is no difference between the two means. The alternative hypothesis (\( H_a: \mu_1 < \mu_2 \) ) shows that supplier 2 is expected to provide greater impact strength. The test, by calculating a t-statistic, helps determine whether to reject the null hypothesis.

The result of a Welch's t-test provides insight into the reliability of differences observed in the sample means. We calculate the t-statistic using both sample means, their standard deviations, and sample sizes. Then, using the t-distribution, we decide if the observed differences are statistically significant, reflecting the true difference in the population means.
Confidence Interval
A confidence interval offers a range of plausible values for a population parameter, like the difference between two means. In the context of this exercise, we calculate a confidence interval to estimate the difference in mean impact strength between gears from two suppliers.

To compute the confidence interval, we determine the point estimate, which is simply the difference between the sample means (\( \bar{x}_2 - \bar{x}_1 \) ). Adding and subtracting the margin of error, derived from the critical t-value and the standard error, gives us the interval (\(CI = (\bar{x}_2 - \bar{x}_1) \pm t_{df, \alpha/2} \times \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \) ).

In the provided exercise, this interval ranges from 17.03 to 44.97. It suggests that we can be 95% confident that the true difference in mean impact strengths lies within this range. Identifying this interval helps determine if the difference is statistically significant and if specific values, like 25, fall within this scope.
Hypothesis Testing
Hypothesis testing is a statistical process used to make inferential decisions about a population based on sample data. It involves establishing a null hypothesis (\( H_0 \) ) which represents no effect or difference and an alternative hypothesis (\( H_a \) ) suggesting the existence of an effect or difference.

In the exercise, two separate hypothesis tests are conducted. Part (a) involves testing whether supplier 2's gears have higher mean impact strength. The second test (part b) examines if the mean difference is at least 25 foot-pounds. For both, we set up respective hypotheses, calculate a t-statistic, and compare it with critical values to decide whether to reject the null.

Failing to reject the null in part b means insufficient evidence supports the claim that the mean difference is at least 25. The calculated p-value also supports this conclusion. Hypothesis testing, through these procedures, aids in substantiating or undermining claims about population parameters based on sample data.
Impact Strength
Impact strength is a vital property in materials, indicating the energy a material can absorb before failure. It is measured in foot-pounds or joules and is crucial in assessing the durability of components like gears.

In the given exercise, impact strength is the key characteristic of plastic gears produced by two suppliers. It relates to their performance in practical applications—gears with higher impact strength can withstand increased stress or load.

The analysis of the gear's impact strength through statistical evaluation determines not only the mean value but also the comparative quality between different suppliers. By using statistical tests, we can assess whether the differences in impact strength are due to materials' properties rather than chance. This kind of analysis helps in quality control and decision-making on supplier selections.

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Most popular questions from this chapter

A paper in Quality Engineering [2013, Vol. 25(1)\(]\) presented data on cycles to failure of solder joints at different temperatures for different types of printed circuit boards (PCB). Failure data for two temperatures \(\left(20\right.\) and \(60^{\circ} \mathrm{C}\) ) for a coppernickel-gold PCB follow. $$\begin{array}{ll}20^{\circ} \mathrm{C} & 218,265,279,282,336,469,496,507,685,685 \\\60^{\circ} \mathrm{C} & 185,242,254,280,305,353,381,504,556,697\end{array}$$ a. Test the null hypothesis at \(\alpha=0.05\) that the cycles to failure are the same at both temperatures. Is the alternative one or two sided? b. Find a \(95 \%\) confidence interval for the difference in the mean cycles to failure for the two temperatures. c. Is the value zero contained in the \(95 \%\) confidence interval? Explain the connection with the conclusion you reached in part (a). d. Do normal probability plots of part cycles to failure indicate any violations of the assumptions for the tests and confidence interval that you performed?

The overall distance traveled by a golf ball is tested by hitting the ball with Iron Byron, a mechanical golfer with a swing that is said to emulate the distance hit by the legendary champion Byron Nelson. Ten randomly selected balls of two different brands are tested and the overall distance measured. $$\begin{aligned}&\text { The data follow: }\\\&\text { Brand } 1: 275,286,287,271,283,271,279,275,263,267\\\&\text { Brand 2: } 258,244,260,265,273,281,271,270,263,268\end{aligned}$$ a. Is there evidence that overall distance is approximately normally distributed? Is an assumption of equal variances justified? b. Test the hypothesis that both brands of ball have equal mean overall distance. Use \(\alpha=0.05 .\) What is the \(P\) -value? c. Construct a \(95 \%\) two-sided \(\mathrm{CI}\) on the mean difference in overall distance for the two brands of golf balls. d. What is the power of the statistical test in part (b) to detect a true difference in mean overall distance of 5 yards? e. What sample size would be required to detect a true difference in mean overall distance of 3 yards with power of approximately \(0.75 ?\)

A photoconductor film is manufactured at a nominal thickness of 25 mils. The product engineer wishes to increase the mean speed of the film and believes that this can be achieved by reducing the thickness of the film to 20 mils. Eight samples of each film thickness are manufactured in a pilot production process, and the film speed (in microjoules per square inch) is measured. For the 25 -mil film, the sample data result is \(\bar{x}_{1}=1.15\) and \(s_{1}=0.11,\) and for the \(20-\mathrm{mil}\) film the data yield \(\bar{x}_{2}=1.06\) and \(s_{2}=0.09 .\) Note that an increase in film speed would lower the value of the observation in microjoules per square inch. a. Do the data support the claim that reducing the film thickness increases the mean speed of the film? Use \(\sigma=0.10\), and assume that the two population variances are equal and the underlying population of film speed is normally distributed. What is the \(P\) -value for this test? b. Find a \(95 \%\) confidence interval on the difference in the two means that can be used to test the claim in part (a).

The New England Journal of Medicine reported an experiment to judge the efficacy of surgery on men diagnosed with prostate cancer. The randomly assigned half of 695(347) men in the study had surgery, and 18 of them eventually died of prostate cancer compared with 31 of the 348 who did not have surgery. Is there any evidence to suggest that the surgery lowered the proportion of those who died of prostate cancer?

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