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Chapter 8: Problem 29

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 mean tire life is [58,200.13, 62,079.27] km.

Step by step solution

01

Understand the Problem

We are asked to find a 95% confidence interval for the mean tire life of a new rubber compound based on a sample of 16 tires. We are given the sample mean (\(\bar{x} = 60,139.7\) km) and the sample standard deviation (\(s = 3645.94\) km).
02

Identify the Correct Formula

Since we have a small sample size (n = 16), we use the t-distribution to find the confidence interval. The formula for a confidence interval using the sample mean is: \[ \bar{x} \pm t_{\alpha/2} \frac{s}{\sqrt{n}} \] where \(t_{\alpha/2}\) is the t-score for \(\alpha/2\) with \(n-1\) degrees of freedom.
03

Determine the t-Score

For a 95% confidence interval with 15 degrees of freedom (n-1 = 15), we look up the t-score. From a t-distribution table or calculator, \(t_{0.025, 15} \approx 2.131.\)
04

Calculate the Margin of Error

The margin of error (ME) is calculated as: \[ ME = t_{\alpha/2} \cdot \frac{s}{\sqrt{n}} = 2.131 \cdot \frac{3645.94}{\sqrt{16}} = 1939.57. \]
05

Calculate the Confidence Interval

Now, we use the margin of error to find the confidence interval: \[ 60,139.7 \pm 1939.57 \] which means the interval is \([58,200.13, 62,079.27]\) km.

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Key Concepts

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

t-distribution
When we're dealing with small sample sizes, like the 16 tires tested in the exercise, the t-distribution becomes very important. The t-distribution is similar to the normal distribution, or bell curve, but it has thicker tails. These thicker tails help provide better estimates when the sample size is small or when the population standard deviation is unknown. This distribution adjusts for the extra uncertainty.
  • The t-distribution becomes more like the normal distribution as the sample size increases.
  • It takes into account the variability among smaller samples.
  • The shape of the t-distribution depends on degrees of freedom, which is one less than the sample size for a single sample.
In our tire testing scenario, the use of t-distribution provides us with the flexibility needed to properly estimate the confidence interval, even with just 16 tires.
sample mean
The concept of the sample mean is straightforward yet very central to statistics. It is the average of all the data points in a sample, and it's a crucial statistic for making inferences about a population from which the sample is drawn. In the tire lifespan study, the sample mean of 60,139.7 kilometers gives a central value around which the entire data set revolves.
  • Calculating the sample mean involves summing up all observed values and dividing by the total number of observations.
  • It provides a representative figure of the overall data.
  • Sample means can vary from sample to sample, but with larger samples, they tend to be closer to the true population mean.
This sample mean acts as the starting point for constructing the confidence interval, providing a baseline estimate of the tire's life expectancy.
degrees of freedom
Degrees of freedom is a fundamental concept often associated with various statistical calculations, including the t-distribution. It refers to the number of values in a calculation that are free to vary. For the exercise in question, which involves calculating the t-score, the degrees of freedom is the sample size minus one.
  • In this tire life estimation, we have 15 degrees of freedom (16 tires - 1).
  • Degrees of freedom affect the shape of the t-distribution, with fewer degrees leading to thicker tails, indicating more variability.
  • A crucial role of degrees of freedom is helping determine which t-value to use when constructing a confidence interval.
Understanding degrees of freedom is key to selecting the correct t-score for your confidence interval calculation.
margin of error
The margin of error is the term we use to describe the range within which we expect the true population parameter (like the mean tire life) to fall, based on the sample data. It quantifies uncertainty and provides an upper and lower bound across which our results extend, given a certain level of confidence.
  • It's calculated as the product of the t-score and the standard error of the sample mean.
  • In the example for the tire lifespan, the margin of error was calculated to be 1939.57 kilometers.
  • The margin of error, when added to and subtracted from the sample mean, gives the boundaries of our confidence interval.
The margin of error is not just a single number but a crucial component that provides depth to statistical estimates, allowing us to say with 95% confidence how the tire's true mean lifespan might vary.

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Most popular questions from this chapter

A random sample has been taken from a normal distribution and the following confidence intervals constructed using the same data: (38.02,61.98) and (39.95,60.05) (a) What is the value of the sample mean? (b) One of these intervals is a \(95 \% \mathrm{CI}\) and the other is a \(90 \%\) CI. Which one is the \(95 \%\) CI and why?

A rivet is to be inserted into a hole. A random sample of \(n=15\) parts is selected, and the hole diameter is measured. The sample standard deviation of the hole diameter measurements is \(s=0.008\) millimeters. Construct a \(99 \%\) lower confidence bound for \(\sigma^{2}\).

During the 1999 and 2000 baseball seasons, there was much speculation that the unusually large number of home runs hit was due at least in part to a livelier ball. One way to test the "liveliness" of a baseball is to launch the ball at a vertical surface with a known velocity \(V_{L}\) and measure the ratio of the outgoing velocity \(V_{0}\) of the ball to \(V_{L} .\) The ratio \(R=V_{0} / V_{L}\) is called the coefficient of restitution. Following are measurements of the coefficient of restitution for 40 randomly selected baseballs. The balls were thrown from a pitching machine at an oak surface. \(\begin{array}{llllll}0.6248 & 0.6237 & 0.6118 & 0.6159 & 0.6298 & 0.6192 \\\ 0.6520 & 0.6368 & 0.6220 & 0.6151 & 0.6121 & 0.6548 \\ 0.6226 & 0.6280 & 0.6096 & 0.6300 & 0.6107 & 0.6392 \\ 0.6230 & 0.6131 & 0.6223 & 0.6297 & 0.6435 & 0.5978 \\ 0.6351 & 0.6275 & 0.6261 & 0.6262 & 0.6262 & 0.6314 \\\ 0.6128 & 0.6403 & 0.6521 & 0.6049 & 0.6170 & \\ 0.6134 & 0.6310 & 0.6065 & 0.6214 & 0.6141 & \end{array}\) (a) Is there evidence to support the assumption that the coefficient of restitution is normally distributed? (b) Find a \(99 \%\) CI on the mean coefficient of restitution. (c) Find a \(99 \%\) prediction interval on the coefficient of restitution for the next baseball that will be tested. (d) Find an interval that will contain \(99 \%\) of the values of the coefficient of restitution with \(95 \%\) confidence. (e) Explain the difference in the three intervals computed in parts (b), (c), and (d).

The wall thickness of 25 glass 2 -liter bottles was measured by a quality- control engineer. The sample mean was \(\bar{x}=4.05\) millimeters, and the sample standard deviation was \(s=0.08\) millimeter. Find a \(95 \%\) lower confidence bound for mean wall thickness. Interpret the interval obtained.

Determine the \(\chi^{2}\) percentile that is required to construct each of the following CIs: (a) Confidence level \(=95 \%,\) degrees of freedom \(=24,\) onesided (upper) (b) Confidence level \(=99 \%,\) degrees of freedom \(=9,\) one-sided (lower) (c) Confidence level \(=90 \%,\) degrees of freedom \(=19,\) two-sided.
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