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Chapter 8: Problem 27

A research engineer for a tire manufacturer is investigating tire life for a new rubber compound and has built 16 tires and tested them to end-of-life in a road test. The sample mean and standard deviation are 60,139.7 and 3645.94 kilometers. Find a \(95 \%\) confidence interval on mean tire life.

Short Answer

Expert verified
The 95% confidence interval for the mean tire life is (58,198.43, 62,081.97) kilometers.

Step by step solution

01

Identify Known Values

We know the following:- Sample mean \( \bar{x} = 60139.7 \) kilometers- Sample standard deviation \( s = 3645.94 \) kilometers- Sample size \( n = 16 \)- Confidence level = 95%We need to find the confidence interval for the population mean.
02

Determine the Critical Value

Since the sample size is small (\( n = 16 \)), we use the t-distribution. The degrees of freedom are \( n-1 = 15 \). For a 95% confidence interval and 15 degrees of freedom, refer to the t-distribution table to find the critical value \( t^* \). For \(15\) degrees of freedom, \( t^* \approx 2.131 \).
03

Calculate the Standard Error

The standard error of the mean is calculated using the formula:\[ SE = \frac{s}{\sqrt{n}} \]Substitute the known values:\[ SE = \frac{3645.94}{\sqrt{16}} = \frac{3645.94}{4} = 911.485 \text{ kilometers} \]
04

Compute the Margin of Error

The margin of error (ME) can be calculated using the critical value and the standard error:\[ ME = t^* \times SE \]\[ ME = 2.131 \times 911.485 \approx 1941.27 \text{ kilometers} \]
05

Construct the Confidence Interval

The confidence interval is calculated using the formula:\[ (\bar{x} - ME, \bar{x} + ME) \]Substitute the known numbers:\[ (60139.7 - 1941.27, 60139.7 + 1941.27) \]\[ (58198.43, 62081.97) \text{ kilometers} \]
06

Interpret the Results

The 95% confidence interval for the mean tire life of the new compound is approximately (58,198.43, 62,081.97) kilometers.

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Key Concepts

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

t-distribution
When a sample size is small or the population standard deviation is unknown, statisticians often use the t-distribution. It is similar to the normal distribution but has heavier tails. This means it is more spread out. The t-distribution allows for more variability, which is useful in small sample sizes.
  • This distribution is symmetric and bell-shaped, like the normal distribution.
  • It is determined by degrees of freedom, which are calculated as the sample size minus one ( \( n - 1 \)).
In the example problem, we use a sample size of 16, leading to 15 degrees of freedom. This value is crucial for finding the right critical value (\( t^* \)) from the t-distribution table.
Sample Size
The sample size, denoted by \( n \), is the number of observations in a sample. It is a critical factor in statistical analysis, affecting the accuracy of your results. A larger sample size gives more reliable and accurate results, while a small sample size may lead to more variability and uncertainty.
  • A sample size of 16 is considered small, hence the use of t-distribution.
  • Larger sample sizes use the normal distribution, which has less variability.
In our tire example, the sample size affects the degrees of freedom, and subsequently, the choice of using the t-distribution for calculations.
Standard Error
The standard error (SE) measures how much the sample mean would differ from the true population mean if you took several samples. It provides an understanding of the accuracy of the sample mean.
  • Calculated by dividing the sample standard deviation by the square root of the sample size ( \( SE = \frac{s}{\sqrt{n}} \)).
  • In general, a smaller SE suggests that the sample mean is a more accurate reflection of the population mean.
In our exercise, the SE was found to be 911.485 kilometers. This indicates the amount of variation in the mean tire life estimate for the given sample.
Margin of Error
The margin of error (ME) gives an interval around the sample mean, indicating the range within which the true population mean is likely to lie with a certain level of confidence. It's critical in constructing confidence intervals.
  • Calculated by multiplying the critical value (\( t^* \)) by the standard error (\( ME = t^* \times SE \)).
  • A larger margin of error indicates less precise estimates of the population parameter.
In the tire life exercise, the margin of error was approximately 1941.27 kilometers. This value helps determine the confidence interval, providing a range that likely contains the actual mean tire life for the new rubber compound.

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Most popular questions from this chapter

A manufacturer of electronic calculators takes a random sample of 1200 calculators and finds that there are eight defective units. (a) Construct a \(95 \%\) confidence interval on the population proportion. (b) Is there evidence to support a claim that the fraction of defective units produced is \(1 \%\) or less?

An article in the Journal of the American Statistical Association (1990, Vol. 85, pp. 972-985) measured the weight of 30 rats under experiment controls. Suppose that there are 12 underweight rats. (a) Calculate a \(95 \%\) two-sided confidence interval on the true proportion of rats that would show underweight from the experiment. (b) Using the point estimate of \(p\) obtained from the preliminary sample, what sample size is needed to be \(95 \%\) confident that the error in estimating the true value of \(p\) is less than \(0.02 ?\) (c) How large must the sample be if we wish to be at least \(95 \%\) confident that the error in estimating \(p\) is less than 0.02, regardless of the true value of \(p ?\)

An article in Urban Ecosystems, "Urbanization and Warming of Phoenix (Arizona, USA): Impacts, Feedbacks and Mitigation" (2002, Vol. 6, pp. \(183-203\) ), mentions that Phoenix is ideal to study the effects of an urban heat island because it has grown from a population of 300,000 to approximately 3 million over the last 50 years and this is a period with a continuous, detailed climate record. The 50 -year averages of the mean annual temperatures at eight sites in Phoenix are shown below. Check the assumption of normality in the population with a probability plot. Construct a \(95 \%\) confidence interval for the standard deviation over the sites of the mean annual temperatures. $$ \begin{array}{lc} \text { Site } & \begin{array}{c} \text { Average Mean } \\ \text { Temperature }\left({ }^{\circ} \mathbf{C}\right) \end{array} \\ \hline \text { Sky Harbor Airport } & 23.3 \\ \text { Phoenix Greenway } & 21.7 \\ \text { Phoenix Encanto } & 21.6 \\ \text { Waddell } & 21.7 \\ \text { Litchfield } & 21.3 \\ \text { Laveen } & 20.7 \\ \text { Maricopa } & 20.9 \\ \text { Harlquahala } & 20.1 \\ \hline \end{array} $$

An article in the Journal of Human Nutrition and Dietetics ["The Validation of Energy and Protein Intakes by Doubly Labeled Water and 24-Hour Urinary Nitrogen Excretion in Post-Obese Subjects" (1995, Vol. 8, pp. \(51-64\) ) ] showed the energy intake expressed as a basal metabolic rate, BMR (MJ). $$ \begin{array}{lllll} 5.40 & 5.67 & 5.79 & 6.85 & 6.92 \end{array} $$ \(\begin{array}{lllll}5.70 & 6.08 & 5.48 & 5.44 & 5.51\end{array}\) (a) Use a normal probability plot to check the normality assumption. (b) Find a \(99 \%\) two-sided confidence interval on the mean BMR

A normal population has known mean \(\mu=50\) and variance \(\sigma^{2}=5\). What is the approximate probability that the sample variance is greater than or equal to \(7.44 ?\) less than or equal to \(2.56 ?\) For a random sample of size (a) \(n=16\) (b) \(n=30\) (c) \(n=71\) (d) Compare your answers to parts (a)-(c) for the approximate probability that the sample variance is greater than or equal to \(7.44 .\) Explain why this tail probability is increasing or decreasing with increased sample size. (e) Compare your answers to parts (a)-(c) for the approximate probability that the sample variance is less than or equal to \(2.56 .\) Explain why this tail probability is increasing or decreasing with increased sample size.
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