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Chapter 10: Problem 39

The manager of a fleet of automobiles is testing two brands of radial tires. He assigns one tire of each brand at random to the two rear wheels of eight cars and runs the cars until the tires wear out. The data (in kilometers) follow. Find a \(99 \%\) confidence interval on the difference in mean life. Which brand would you prefer, based on this calculation? $$\begin{array}{ccc}\hline \text { Car } & \text { Brand 1 } & \text { Brand 2 } \\ \hline 1 & 36,925 & 34,318 \\ 2 & 45,300 & 42,280 \\ 3 & 36,240 & 35,500 \\ 4 & 32,100 & 31,950 \\ 5 & 37,210 & 38,015 \\ 6 & 48,360 & 47,800 \\\ 7 & 38,200 & 37,810 \\ 8 & 33,500 & 33,215 \\ \hline \end{array}$$

Short Answer

Expert verified
There is no significant difference between the two brands; I have no preference based on this analysis.

Step by step solution

01

Calculate the Differences

First, calculate the difference in the kilometers for each car (Brand 1 - Brand 2). This will give us the differences: 1. 36,925 - 34,318 = 2,607 2. 45,300 - 42,280 = 3,020 3. 36,240 - 35,500 = 740 4. 32,100 - 31,950 = 150 5. 37,210 - 38,015 = -805 6. 48,360 - 47,800 = 560 7. 38,200 - 37,810 = 390 8. 33,500 - 33,215 = 285.
02

Mean of the Differences

Calculate the mean of the differences. \[\bar{x} = \frac{1}{8}(2607 + 3020 + 740 + 150 - 805 + 560 + 390 + 285) = 1231.875\]
03

Calculate the Standard Deviation of Differences

Find the standard deviation of the differences. First, calculate each deviation squared and then find the average.Find the squared deviations:\((2607 - 1231.875)^2 = 1896653.52\)\((3020 - 1231.875)^2 = 3200740.52\)\((740 - 1231.875)^2 = 241031.64\)\((150 - 1231.875)^2 = 1168895.02\)\((-805 - 1231.875)^2 = 4140139.02\)\((560 - 1231.875)^2 = 450714.14\)\((390 - 1231.875)^2 = 708985.52\)\((285 - 1231.875)^2 = 894644.14\)The sum of squared deviations is:\[ \sum (x_i - \bar{x})^2 = 14493803.48 \]Calculate the variance:\[ s^2 = \frac{14493803.48}{7} \approx 2070543.354\]Calculate the standard deviation:\[ s = \sqrt{2070543.354} \approx 1438.10 \]
04

Determine the t-Value for 99% CI

For 99% confidence level and 7 degrees of freedom (since sample size is 8, df = n-1 = 7), we find the t-value from the t-table, which is approximately 3.499.
05

Calculate the Margin of Error

Calculate the margin of error (ME):\[ME = t \times \frac{s}{\sqrt{n}} = 3.499 \times \frac{1438.10}{\sqrt{8}}\]\[ME = 3.499 \times 508.21 \approx 1777.54\]
06

Calculate the Confidence Interval

The confidence interval is calculated as follows:\[(\bar{x} - ME, \bar{x} + ME) = (1231.875 - 1777.54, 1231.875 + 1777.54) = (-545.665, 3009.415)\]
07

Interpretation of Results

The 99% confidence interval for the difference in mean tire life is (-545.665, 3009.415). Since the interval includes zero, there is not enough evidence to suggest a significant difference between the two brands at the 99% confidence level.

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Key Concepts

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

Confidence Interval
A confidence interval is a range of values used to estimate the true value of a population parameter, like the mean. This range is constructed from sample data and is accompanied by a confidence level, which indicates how sure we are that the interval contains the true parameter value. For instance, a 99% confidence interval suggests that if we took many samples and built an interval from each, approximately 99% of those intervals would contain the true mean.

To create a confidence interval, we use the sample mean as a starting point. Adding and subtracting a margin of error to this mean gives us the interval. The margin of error depends on the standard deviation of the sample and the t-value associated with our desired confidence level, as demonstrated in the solution above.

In the tire example, the 99% confidence interval for the difference in mean tire life gives a range of values within which we expect the true difference to fall. Because zero is included in this interval, it is not significantly different at the 99% confidence level. This means that brands may not differ in performance.
Hypothesis Testing
Hypothesis testing is a statistical method used to make decisions based on data analysis. It starts with a null hypothesis, which is a statement that there is no effect or no difference. In contrast, the alternative hypothesis suggests there is a difference or an effect.

Using the tire example, the null hypothesis ( H_0 ) could be that there is no difference in the mean lifespan of the two tire brands. The alternative hypothesis ( H_a ) proposes that there is a difference. By comparing confidence intervals, we test the null hypothesis.

The calculated confidence interval in this example did not provide sufficient evidence to reject the null hypothesis. This is because the interval encompasses zero, meaning any observed difference could just be due to random variation in the sample data rather than a real difference between the two tire brands.
  • Null Hypothesis ( H_0 ): No difference between means.
  • Alternative Hypothesis ( H_a ): A difference exists.
  • Decision based on whether zero lies within the confidence interval.
Comparison of Means
Comparison of means is a statistical analysis used to determine if there is a significant difference between two or more group means. In our tire brands example, we compare the means of the kilometers run by each brand's tires.

When comparing two means, as we did with the tire brands, the main aim is to see if one mean is consistently higher or lower than the other. We achieve this by looking at the difference in means and calculating a confidence interval around this difference. If the interval does not include zero, we might conclude that the means are different.

Here, even though one brand had a slightly higher mean tire life, the interval had zero within it, indicating no statistically significant difference at the 99% confidence level.
  • Comparison based on the calculated difference of sample means.
  • Interval analysis helps determine if the observed difference is statistically significant.
  • In the study, the result was inconclusive due to the interval containing zero.

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Most popular questions from this chapter

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: $$\begin{aligned} \text { Catalyst } 1: & 57.9,66.2,65.4,65.4,65.2,62.6,67.6,63.7, 67.2,71.0\end{aligned}$$ $$\begin{array}{l} \text { Catalyst 2: } 66.4,71.7,70.3,69.3 ,64.8,69.6,68.6,69.4,65.3,68.8\end{array}$$ (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?

Consider the hypothesis test \(H_{0}: \mu_{1}=\mu_{2}\) against \(H_{1}: \mu_{1}>\mu_{2} .\) Suppose that sample sizes \(n_{1}=10\) and \(n_{2}=10\) that \(\bar{x}_{1}=7.8\) and \(\bar{x}_{2}=5.6,\) and that \(s_{1}^{2}=4\) and \(s_{2}^{2}=9\). 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) if \(\mu_{1}\) is 3 units greater than \(\mu_{2}\)? (d) Assuming equal sample sizes, what sample size should be used to obtain \(\beta=0.05\) if \(\mu_{1}\) is 3 units greater than \(\mu_{2}\) ? Assume that \(\alpha=0.05\).

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.

The thickness of a plastic film (in mils) on a substrate material is thought to be influenced by the temperature at which the coating is applied. A completely randomized experiment is carried out. Eleven substrates are coated at \(125^{\circ} \mathrm{F}\), resulting in a sample mean coating thickness of \(\bar{x}_{1}=103.5\) and a sample standard deviation of \(s_{1}=10.2\). Another 13 substrates are coated at \(150^{\circ} \mathrm{F}\), for which \(\bar{x}_{2}=99.7\) and \(s_{2}=20.1\) are observed. It was originally suspected that raising the process temperature would reduce mean coating thickness. (a) Do the data support this claim? Use \(\alpha=0.01\) and assume that the two population standard deviations are not equal. Calculate an approximate \(P\) -value for this test. (b) How could you have answered the question posed regarding the effect of temperature on coating thickness by using a confidence interval? Explain your answer.

A random sample of 500 adult residents of Maricopa County found that 385 were in favor of increasing the highway speed limit to \(75 \mathrm{mph}\), while another sample of 400 adult residents of Pima County found that 267 were in favor of the increased speed limit. (a) Do these data indicate that there is a difference in the support for increasing the speed limit between the residents of the two counties? Use \(\alpha=0.05 .\) What is the \(P\) -value for this test? (b) Construct a \(95 \%\) confidence interval on the difference in the two proportions. Provide a practical interpretation of this interval.
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