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To test a new tread design with respect to stopping distance, a tire manufacturer manufactures a set of prototype tires and measures the stopping distance from 70 mph on a standard test car. A sample of 25 stopping distances yielded a sample mean 173 feet with sample standard deviation 8 feet. Construct a \(98 \%\) confidence interval for the mean stopping distance for these tires. Assume a normal distribution of stopping distances.

Short Answer

Expert verified
The 98% confidence interval for the mean stopping distance is (168.52, 177.48) feet.

Step by step solution

01

Identify the Given Information

The problem provides us with a sample of 25 stopping distances. The sample mean is \( \bar{x} = 173 \) feet, the sample standard deviation is \( s = 8 \) feet, and the sample size \( n = 25 \). We are asked to compute a \( 98\% \) confidence interval for the population mean.
02

Determine the Critical Value

Since the sample size is less than 30, we use the \( t \)-distribution to find the critical value. With \( n - 1 = 24 \) degrees of freedom and a confidence level of \( 98\% \), we look up a \( t \)-value from the \( t \)-distribution table. This critical value is approximately \( t^* = 2.797 \).
03

Compute the Standard Error of the Mean

The standard error of the mean (SEM) is calculated using the formula: \[ \text{SEM} = \frac{s}{\sqrt{n}} \] Substituting in the values gives: \[ \text{SEM} = \frac{8}{\sqrt{25}} = \frac{8}{5} = 1.6 \] feet.
04

Calculate the Confidence Interval

The confidence interval is calculated using the formula: \[ \bar{x} \pm t^* \times \text{SEM} \] Substituting the known values, we get: \[ 173 \pm 2.797 \times 1.6 \] Calculating the margin of error: \[ 2.797 \times 1.6 = 4.4752 \] Thus, the confidence interval is: \[ 173 \pm 4.4752 = (168.5248, 177.4752) \] feet.

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Key Concepts

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

t-distribution
The t-distribution is crucial when you're dealing with smaller sample sizes and unknown population standard deviations. Unlike the normal distribution, which is symmetric and bell-shaped, the t-distribution has thicker tails. These tails account for the increased variability and uncertainty that comes with small samples. As the sample size increases, the t-distribution approaches the standard normal distribution.
For confidence intervals involving a sample size less than 30, like in our tire test example, the t-distribution is the tool of choice. You use it to find the critical t-value, which depends on the degrees of freedom. The degrees of freedom is generally the sample size minus one (n - 1). In our case, with 25 samples, we have 24 degrees of freedom. This t-value from the t-distribution table is then used to construct the confidence interval around the sample mean.
standard error
The standard error of the mean (SEM) helps estimate the precision of the sample mean as an estimate of the population mean. Think of SEM as a measure of how far the sample mean of your data is likely to be from the true population mean.
The formula for SEM is:
  • \[ \text{SEM} = \frac{s}{\sqrt{n}} \]where \( s \) is the sample standard deviation, and \( n \) is the sample size.
In the tire stopping distance test, the sample standard deviation was 8 feet and the sample size was 25, resulting in a SEM of 1.6 feet. This SEM is then used with the t-value to calculate the confidence interval. A smaller SEM indicates more precise estimates of the population mean.
sample mean
The sample mean, represented by \( \bar{x} \), is an average of the sample data collected. It is a good estimator of the population mean when the data is normally distributed. To calculate the sample mean, sum all observations and divide by the number of observations.
In our example, the stopping distances of the tires were recorded, resulting in a sample mean of 173 feet. This figure is at the center of our confidence interval calculation.
The sample mean is particularly important because it serves as the basis for constructing the confidence interval. By understanding how close the sample mean is to the expected population mean, you can refine the expected range of outcomes (confidence interval) for similar tests.

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Most popular questions from this chapter

Estimate the minimum sample size needed to form a confidence interval for the proportion of a population that has a particular characteristic, meeting the criteria given. a. \(95 \%\) confidence, \(E=0.02\) b. \(99 \%\) confidence, \(E=0.02\) c. \(95 \%\) confidence, \(E=0.01\)

The designer of a garbage truck that lifts roll-out containers must estimate the mean weight the truck will lift at each collection point. A random sample of 325 containers of garbage on current collection routes yielded \(x-=75.3 \mathrm{lb}, \mathrm{s}=12.8 \mathrm{lb}\). Construct a \(99.8 \%\) confidence interval for the mean weight the trucks must lift each time.

Confidence intervals constructed using the formula in this section often do not do as well as expected unless \(n\) is quite large, especially when the true population proportion is close to either 0 or \(1 .\) In such cases a better result is obtained by adding two successes and two failures to the actual data and then computing the confidence interval. This is the same as using the formula $$ \begin{aligned} &\tilde{p} \pm z_{\alpha} / 2 \sqrt{\frac{\hat{p}(1-\dot{p})}{\tilde{n}}}\\\ &\text { where }\\\ &\tilde{p}-\frac{x+2}{n+4} \operatorname{and} n-n+4 \end{aligned} $$ Suppose that in a random sample of 600 households, 12 had no telephone service of any kind. Use the adjusted confidence interval procedure just described to form a \(99.9 \%\) confidence interval for the proportion of all households that have no telephone service of any kind.

A government agency wishes to estimate the proportion of drivers aged \(16-24\) who have been involved in a traffic accident in the last year. It wishes to make the estimate to within one percentage point and at \(90 \%\) confidence. Find the minimum sample size required, using the information that several years ago the proportion was 0.12 .

The number of trips to a grocery store per week was recorded for a randomly selected collection of households, with the results shown in the table. $$ \begin{array}{lllllllll} 2 & 2 & 2 & 1 & 4 & 2 & 3 & 2 & 5 & 4 \\ 2 & 3 & 5 & 0 & 3 & 2 & 3 & 1 & 4 & 3 \\ 3 & 2 & 1 & 6 & 2 & 3 & 3 & 2 & 4 & 4 \end{array} $$ Construct a \(95 \%\) confidence interval for the average number of trips to a grocery store per week of all households.

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