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Chapter 8: Problem 118

Owing to the variability of trade-in allowance, the profit per new car sold by an automobile dealer varies from car to car. The profits per sale (in hundreds of dollars), tabulated for the past week, were \(2.1,3.0,1.2,6.2,4.5,\) and \(5.1 .\) Find a \(90 \%\) confidence interval for the mean profit per sale. What assumptions must be valid for the technique that you used to be appropriate?

Short Answer

Expert verified
The 90% confidence interval is approximately (2.24, 5.12). Assumes normal distribution or sufficient sample size.

Step by step solution

01

Calculate the Sample Mean

To find the sample mean \( \bar{x} \), add all the given profits and divide by the number of data points. Thus, \( \bar{x} = \frac{2.1+3.0+1.2+6.2+4.5+5.1}{6} = \frac{22.1}{6} = 3.6833 \).
02

Calculate the Sample Standard Deviation

The sample standard deviation \( s \) is needed to measure the spread of profits. Calculate \( s \) using the formula \( s = \sqrt{\frac{\sum{(x_i - \bar{x})^2}}{n-1}} \), where \( n \) is the number of samples. For our data: \( s = \sqrt{\frac{(2.1-3.6833)^2 + (3.0-3.6833)^2 + (1.2-3.6833)^2 + (6.2-3.6833)^2 + (4.5-3.6833)^2 + (5.1-3.6833)^2}{5}} \). Calculating this gives \( s \approx 1.752 \).
03

Find the T-Score for Confidence Level

For a \( 90\% \) confidence interval and \( 5 \) degrees of freedom (\( n-1 \)), the critical value from the t-distribution table is \( t^* \approx 2.0158 \).
04

Calculate the Margin of Error

The margin of error (ME) is calculated using \( ME = t^* \cdot \frac{s}{\sqrt{n}} \). Substituting the values gives \( ME = 2.0158 \times \frac{1.752}{\sqrt{6}} \approx 1.44 \).
05

Construct the Confidence Interval

The \( 90\% \) confidence interval is given by \( \bar{x} \pm ME \). Substitute the mean and margin of error to get \( 3.6833 \pm 1.44 \). Thus, the confidence interval is approximately \( (2.2433, 5.1233) \).
06

State Assumptions

The assumptions for using the t-distribution are that the sample is randomly selected from a normal distribution, or the sample size is sufficiently large for the Central Limit Theorem to apply. In small samples like this one, we assume normality.

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Key Concepts

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

Sample Mean
The sample mean is a fundamental concept in statistics that represents the average of a set of values. To find the sample mean, you sum up all the data points and then divide by the number of data points in the sample. It serves as an estimate of the population mean when the entire population cannot be measured due to time or cost constraints.

Here's how it works:
  • Add up all the numbers in your sample.
  • Count how many numbers you have.
  • Divide the total sum by the count.
For instance, in the exercise given, the sample mean of profits from the car sales was calculated as:\[\bar{x} = \frac{2.1+3.0+1.2+6.2+4.5+5.1}{6} = 3.6833\]Remember, the sample mean is different from the population mean because it only considers the sample data you have, not the entire set of possible data.
Sample Standard Deviation
The sample standard deviation provides a measure of variability or spread in a set of data. It tells us how much the individual data points deviate from the sample mean, offering insight into the distribution's consistency.

The calculation involves a few steps:
  • Find the difference between each data point and the sample mean.
  • Square these differences to make them positive.
  • Sum all the squared differences.
  • Divide by the sample size minus one (this is known as Bessel's correction).
  • Take the square root of that division result to get the standard deviation.
For the given exercise, it was calculated as approximately:\[s = \sqrt{\frac{(2.1-3.6833)^2 + (3.0-3.6833)^2 + ... + (5.1-3.6833)^2}{5}} \approx 1.752\]Understanding the sample standard deviation helps you grasp how much variation there is from the average, which is crucial for making predictions and decisions.
T-Distribution
The t-distribution is a crucial tool in statistics used when the sample size is small, and the population standard deviation is unknown. It resembles the normal distribution but has heavier tails, providing more flexibility in statistical analyses.

Key characteristics of the t-distribution include:
  • Dependent on degrees of freedom, which in turn equals the sample size minus one \( (n-1) \).
  • As sample size increases, the t-distribution approaches a normal distribution.
  • Used primarily for hypothesis testing and constructing confidence intervals.
In the exercise, we used the t-distribution to find the critical t-score for a 90% confidence level and 5 degrees of freedom. This t-score helps us build a confidence interval for our sample mean.
Margin of Error
The margin of error is an essential measure in statistics that quantifies the range of values within which the true population parameter lies, with a specified level of confidence. It's an indication of the precision of a sample mean as an estimate of the population mean.

To compute the margin of error:
  • Determine the critical value from a t-distribution based on your desired confidence level and the degrees of freedom.
  • Multiply this critical value by the sample standard deviation divided by the square root of the sample size.
In our exercise, the margin of error was determined to be:\[ME = t^* \cdot \frac{s}{\sqrt{n}} = 2.0158 \times \frac{1.752}{\sqrt{6}} \approx 1.44\]The margin of error creates an interval around the sample mean. This interval provides a range that is likely to include the true population mean with a specified probability, such as 90%.

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Most popular questions from this chapter

We noted in Section 8.3 that if $$S^{\prime 2}=\frac{\sum_{i=1}^{n}\left(Y_{i}-\bar{Y}\right)^{2}}{n} \text { and } S^{2}=\frac{\sum_{i=1}^{n}\left(Y_{i}-\bar{Y}\right)^{2}}{n-1}$$ then \(S^{\prime 2}\) is a biased estimator of \(\sigma^{2}\), but \(S^{2}\) is an unbiased estimator of the same parameter. If we sample from a normal population, a. find \(V\left(S^{\prime 2}\right)\) b. show that \(V\left(S^{2}\right)>V\left(S^{\prime 2}\right)\)

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Scholastic Assessment Test (SAT) scores, which have fallen slowly since the inception of the test, have now begun to rise. Originally, a score of 500 was intended to be average. The mean scores for 2005 were approximately 508 for the verbal test and 520 for the mathematics test. A random sample of the test scores of 20 seniors from a large urban high school produced the means and standard deviations listed in the accompanying table: $$\begin{array}{lcc} \hline & \text { Verbal } & \text { Mathematics } \\ \hline \text { Sample mean } & 505 & 495 \\ \text { Sample standard deviation } & 57 & 69 \\ \hline \end{array}$$ a. Find a \(90 \%\) confidence interval for the mean verbal SAT scores for high school seniors from the urban high school. b. Does the interval that you found in part (a) include the value 508 , the true mean verbal SAT score for \(2005 ?\) What can you conclude? c. Construct a \(90 \%\) confidence interval for the mean mathematics SAT score for the urban high school seniors. Does the interval include 520 , the true mean mathematics score for \(2005 ?\) What can you conclude?

If \(Y_{1}, Y_{2}, \ldots, Y_{n}\) denote a random sample from an exponential distribution with mean \(\theta\), then \(E\left(Y_{i}\right)=\theta\) and \(V\left(Y_{i}\right)=\theta^{2} .\) Thus, \(E(\bar{Y})=\theta\) and \(V(\bar{Y})=\theta^{2} / n,\) or \(\sigma_{\bar{Y}}=\theta / \sqrt{n} .\) Suggest an unbiased estimator for \(\theta\) and provide an estimate for the standard error of your estimator.

If \(Y\) has a binomial distribution with parameters \(n\) and \(p\), then \(\hat{p}_{1}=Y / n\) is an unbiased estimator of \(p .\) Another estimator of \(p\) is \(\hat{p}_{2}=(Y+1) /(n+2)\) a. Derive the bias of \(\hat{p}_{2}\) b. Derive \(\operatorname{MSE}\left(\hat{p}_{1}\right)\) and \(\operatorname{MSE}\left(\hat{p}_{2}\right)\) c. For what values of \(p\) is \(\operatorname{MSE}\left(\hat{p}_{1}\right)<\operatorname{MSE}\left(\hat{p}_{2}\right) ?\)
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