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Chapter 10: Problem 20

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 \(s_{1}=12\), while 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
(a) Reject \(H_0\), strong evidence supplier 2 is better. (b) Fail to reject \(H_0\), no evidence for 25 difference. (c) CI: (16.9, 45.1), does not fully support 25 higher claim.

Step by step solution

01

State Hypotheses for Part (a)

For part (a), we need to test if supplier 2 has a higher mean impact strength than supplier 1. We use the following hypotheses:- Null hypothesis: \( H_0: \mu_1 = \mu_2 \)- Alternative hypothesis: \( H_a: \mu_1 < \mu_2 \)
02

Determine Test Statistic for Part (a)

Since the standard deviations are unequal, we'll use the two-sample t-test for means with unequal variances. The test statistic is given by:\[ t = \frac{\bar{x}_2 - \bar{x}_1}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} = \frac{321 - 290}{\sqrt{\frac{12^2}{10} + \frac{22^2}{16}}}\]Calculate \( t \) using the given data.
03

Calculate Test Statistic for Part (a)

Substitute the values to calculate:\[ t = \frac{31}{\sqrt{\frac{144}{10} + \frac{484}{16}}} \approx \frac{31}{\sqrt{14.4 + 30.25}} \approx \frac{31}{\sqrt{44.65}} \approx \frac{31}{6.68} \approx 4.64\]Thus, \( t \approx 4.64 \).
04

Determine P-value and Decision for Part (a)

With \( t \approx 4.64 \), use a t-distribution table or calculator with degrees of freedom approximated by:\[ df \approx \frac{\left( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right)^2}{\frac{\left( \frac{s_1^2}{n_1} \right)^2}{n_1 - 1} + \frac{\left( \frac{s_2^2}{n_2} \right)^2}{n_2 - 1}} \approx 17.68\]The \( p \)-value for \( t = 4.64 \) with 17-18 degrees of freedom is much smaller than 0.05, indicating strong evidence against \( H_0 \). We reject \( H_0 \), supporting the claim that supplier 2 provides higher mean impact strength.
05

State Hypotheses for Part (b)

For part (b), test the claim that the mean impact strength of supplier 2 is at least 25 foot-pounds higher than that of supplier 1:- Null hypothesis: \( H_0: \mu_2 - \mu_1 = 25 \)- Alternative hypothesis: \( H_a: \mu_2 - \mu_1 > 25 \)
06

Determine Test Statistic for Part (b)

Use the same test statistic formula as in part (a) but for the new difference:\[ t = \frac{\bar{x}_2 - \bar{x}_1 - 25}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} = \frac{321 - 290 - 25}{6.68}\]Calculate \( t \) using these values.
07

Calculate Test Statistic for Part (b)

\[ t = \frac{6}{6.68} \approx 0.90\]Here, \( t \approx 0.90 \).
08

Determine P-value and Decision for Part (b)

With \( t \approx 0.90 \) and approximately 17-18 degrees of freedom, the \( p \)-value is greater than 0.05. We fail to reject \( H_0 \). The data do not support the claim that the mean impact strength of gears from supplier 2 is at least 25 foot-pounds higher.
09

Construct Confidence Interval for Part (c)

Construct a 95% confidence interval for \( \mu_2 - \mu_1 \) using the formula:\[ (\bar{x}_2 - \bar{x}_1) \pm t_{\alpha/2, df} \times \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}\]Using \( t_{0.025, 17} \approx 2.11 \) for 95% confidence with approximate degrees of freedom.
10

Calculate Confidence Interval for Part (c)

\[ (321 - 290) \pm 2.11 \times 6.68 \approx 31 \pm 14.10\]So, the confidence interval is approximately (16.9, 45.1). This indicates we can expect the true difference in means to fall within this range.
11

Interpret Confidence Interval for Part (c)

The confidence interval (16.9, 45.1) does not exclude zero, so it suggests supplier 2 could have a higher mean. However, it does not support that the mean difference is at least 25 foot-pounds with high confidence, as 25 is not contained within the interval.

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Key Concepts

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

Two-sample t-test
The two-sample t-test is a statistical method used to determine if there is a significant difference between the means of two independent groups. In this scenario with the plastic gears, it helps us compare the impact strengths of gears from two different suppliers. Since the variances are unequal, we use the Welch's t-test, which adjusts the degrees of freedom to account for this disparity.

The procedure involves the following steps:
  • State the null hypothesis ( \(H_0: \mu_1 = \mu_2 \) ) and the alternative hypothesis ( \(H_a: \mu_1 < \mu_2 \) ), implying that supplier 2 has a higher mean strength.
  • Calculate the test statistic using the formula: \[ t = \frac{\bar{x}_2 - \bar{x}_1}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]
  • Determine the P-value using the calculated t-statistic and the degrees of freedom.
  • Compare the P-value with the significance level (\( \alpha =0.05 \) ) to decide whether to reject the null hypothesis.
In our example, the strong evidence against the null hypothesis indicates that supplier 2 indeed provides gears with higher mean impact strength.
Confidence Intervals
A confidence interval provides a range of values that you can be fairly sure your parameter lies within. It gives an estimate of the reliability and precision of your sample data. In the context of comparing two suppliers, a confidence interval for the difference in means tells you how much more, or less, one mean is likely to be compared to another.

To construct a confidence interval, follow these steps:
  • Calculate the difference between sample means \( (\bar{x}_2 - \bar{x}_1) \).
  • Determine the standard error of the difference, factoring in both sample sizes and variances: \[ SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \]
  • Find the appropriate t-value for your confidence level and degrees of freedom.
  • Apply the formula: \[ (\bar{x}_2 - \bar{x}_1) \pm t_{\alpha/2, df} \times SE \]
This interval helps us understand the potential variability in the difference in gear strength between suppliers. Our example interval does not support a significant 25 foot-pound difference, thus providing crucial insight into supplier performance.
Degrees of Freedom
Degrees of freedom are crucial in determining the shape of the t-distribution used in hypothesis testing. In simple terms, they refer to the number of values in a calculation that are free to vary. With unequal variances, the degrees of freedom for the two-sample t-test can be approximated using Welch's method.

The formula is:\[ df \approx \frac{\left( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right)^2}{\frac{\left( \frac{s_1^2}{n_1} \right)^2}{n_1-1} + \frac{\left( \frac{s_2^2}{n_2} \right)^2}{n_2-1}}\]This approximation accounts for differences in sample variances and sizes, providing a more accurate P-value for the hypothesis test.

Understanding degrees of freedom and their impact on statistical tests is vital. It helps ensure that the results are reliable, accounting for the variability in your data. In the gear testing scenario, correctly calculating them informs whether the observed difference between supplier means is genuine.

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Most popular questions from this chapter

Two catalysts may be used in a batch chemical process. Twelve batches were prepared using catalyst 1 , resulting in an average yield of 86 and a sample standard deviation of \(3 .\) Fifteen batches were prepared using catalyst \(2,\) and they resulted in an average yield of 89 with a standard deviation of \(2 .\) Assume that yield measurements are approximately normally distributed with the same standard deviation. (a) Is there evidence to support a claim that catalyst 2 produces a higher mean yield than catalyst \(1 ?\) Use \(\alpha=0.01\). (b) Find a \(99 \%\) confidence interval on the difference in mean yields that can be used to test the claim in part (a).

One of the authors travels regularly to Seattle, Washington. He uses either Delta or Alaska. Flight delays are sometimes unavoidable, but he would be willing to give most of his business to the airline with the best on-time arrival record. The number of minutes that his flight arrived late for the last six trips on each airline follows. Is there evidence that either airline has superior on-time arrival performance? Use \(\alpha=0.01\) and the Wilcoxon rank-sum test. $$ \begin{array}{l|l} \text { Delta: } & 13,10,1,-4, \quad 0,9 \text { (minutes late) } \\ \hline \text { Alaska: } & 15,8,3,-1,-2,4 \text { (minutes late) } \end{array} $$

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 \(\alpha=0.05\).

Two different analytical tests can be used to determine the impurity level in steel alloys. Eight specimens are tested using both procedures, and the results are shown in the following tabulation. (a) Is there sufficient evidence to conclude that tests differ in the mean impurity level, using \(\alpha=0.01 ?\) (b) Is there evidence to support the claim that Test 1 generates a mean difference 0.1 units lower than Test 2 ? Use \(\alpha=0.05\). (c) If the mean from Test 1 is 0.1 less than the mean from Test 2 , it is important to detect this with probability at least \(0.90 .\) Was the use of eight alloys an adequate sample size? If not, how many alloys should have been used?

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 \(s_{1}=0.11,\) while for the 20 -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 \(\alpha=0.10\) 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).
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