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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 is (58,198.51, 62,080.89) kilometers.

Step by step solution

01

Identify the Parameters

Given in the problem, the sample mean \(\bar{x} = 60,139.7\) kilometers and the sample standard deviation \(s = 3,645.94\) kilometers. The sample size \(n = 16\). We need to find a 95% confidence interval for the population mean tire life.
02

Determine the Confidence Level

The confidence level is 95%. This means the level of risk or alpha \(\alpha = 1 - 0.95 = 0.05\). Since this is a two-tailed test, \(\alpha/2 = 0.025\) on each side of the distribution.
03

Find the Critical Value

Since the sample size is small (\(n = 16\) is less than 30), use the t-distribution to find the critical value. The degrees of freedom \(df = n - 1 = 15\). Consult a t-distribution table or use a calculator to find \(t_{0.025, 15} \approx 2.131\) (this is the critical t-value for 15 degrees of freedom at \(\alpha/2 = 0.025\)).
04

Calculate the Standard Error

The standard error (SE) of the sample mean is calculated using the formula: \[SE = \frac{s}{\sqrt{n}} = \frac{3,645.94}{\sqrt{16}} = 911.485\text{ kilometers.}\]
05

Calculate the Margin of Error

The margin of error (ME) is calculated by multiplying the critical t-value by the standard error:\[ME = t \times SE = 2.131 \times 911.485 = 1,941.192\text{ kilometers.}\]
06

Determine the Confidence Interval

The confidence interval (CI) for the population mean tire life is then:\[CI = \left( \bar{x} - ME, \bar{x} + ME \right) = \left( 60,139.7 - 1,941.192, 60,139.7 + 1,941.192 \right) = (58,198.508, 62,080.892)\text{ kilometers.}\]

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Key Concepts

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

t-distribution
When dealing with small sample sizes, typically less than 30, statisticians prefer using the t-distribution rather than the standard normal distribution (also known as the z-distribution). This is because the t-distribution accounts for the extra variability that often occurs with small samples. The t-distribution is similar in shape to the normal distribution but has thicker tails, meaning there's a higher probability of extreme values. This makes it more reliable for smaller samples.

For our example, since the sample size is 16 (which is less than 30), we apply the t-distribution to find the critical value needed for calculating the confidence interval. The critical value depends on the desired confidence level and the degrees of freedom, which is calculated as the sample size minus one ( - 1).
  • Degrees of freedom (df): In our case: df = 16 - 1 = 15.
  • Confidence level: We chose a 95% confidence level, which means there is a 5% chance of the true mean falling outside this interval (2.5% on each extreme of the distribution).
  • Critical t-value: Using a t-table or calculator, the critical t-value for 15 degrees of freedom at a 95% confidence level is approximately 2.131.
sample mean
The sample mean is the average value of a set of data, and it's a key indicator of the central tendency in your data. It gives us a central value around which our data tends to cluster.

In the tire test example, the sample mean signifies the average tire life measured from the tested sample of tires. Specifically, the sample mean is calculated as follows:
  • Collect all individual tire life measurements obtained from the test.
  • Add these measurements together to get a total sum.
  • Divide this sum by the sample size, which is the total number of tires tested. In this case, that number is 16.
For our exercise, the sample mean calculated was 60,139.7 kilometers. This value serves as the basis for further calculations, helping us estimate the range in which the true average tire life of all manufactured tires is likely to fall.
standard error
The standard error (SE) is an important statistic that measures how much the sample mean is expected to vary from the true population mean. It's essentially a gauge of the accuracy of our sample mean in reflecting the true population mean.

To calculate the standard error, you need the sample's standard deviation and the sample size. The formula used is as follows:\[SE = \frac{s}{\sqrt{n}}\]where:
  • s: is the sample standard deviation. In our example, it is 3,645.94 kilometers.
  • n: is the sample size, which is 16 for the tire test.
So, for the tire life example:
  • The standard error calculated is \( SE = \frac{3,645.94}{\sqrt{16}} = 911.485 \) kilometers.
The standard error provides a window into the variability of our sample mean. A smaller standard error signifies that the sample mean is likely a more accurate reflection of the population mean.
margin of error
The margin of error (ME) is a critical component when constructing confidence intervals. It represents the range of values above and below the sample mean that you expect to encompass the true population mean with a certain level of confidence.

To compute the margin of error, we multiply the standard error by the critical t-value obtained from the t-distribution. Here is the formula:\[ME = t \times SE\]where:
  • t: is the critical t-value (2.131 for a 95% confidence level with 15 degrees of freedom).
  • SE: is the standard error (911.485 kilometers in our example).
The margin of error in the tire life example comes out as:
  • \( ME = 2.131 \times 911.485 = 1941.192 \) kilometers.
This margin indicates how much we expect our sample mean to vary from the true population mean, thus allowing us to derive the confidence interval. By adding and subtracting this margin from the sample mean, we can determine the range within which we are 95% confident the true mean tire life falls.

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Most popular questions from this chapter

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