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An automobile manufacturer who wishes to advertise that one of its models achieves \(30 \mathrm{mpg}\) (miles per gallon) decides to carry out a fuel efficiency test. Six nonprofessional drivers are selected, and each one drives a car from Phoenix to Los Angeles. The resulting fuel efficiencies (in miles per gallon) are: \(\begin{array}{llllll}27.2 & 29.3 & 31.2 & 28.4 & 30.3 & 29.6\end{array}\) Assuming that fuel efficiency is normally distributed under these circumstances, do the data contradict the claim that true average fuel efficiency is (at least) \(30 \mathrm{mpg}\) ?

Short Answer

Expert verified
The answer depends on the calculated t-value compared to the critical t-value. If the calculated value is less, the data do not contradict the manufacturer's claim. If the calculated value is greater, the data suggest that the true average fuel efficiency may be less than 30 mpg.

Step by step solution

01

Calculate the Sample Mean and Standard Deviation

First, calculate the mean (average) of the six observed fuel efficiencies by adding all six values and dividing by the number of values, which is 6 in this case. Similarly, calculate the sample standard deviation, which is the square root of the variance. Variance is the average of the squared differences from the mean.
02

Conduct a One-Sample T-test

Next, conduct a one-sample t-test. Assume the null hypothesis is that the true population mean is 30 mpg. Calculate the t-value using the formula \((\bar{x} - \mu) / (\sigma / \sqrt{n})\), where \(\bar{x}\) is the sample mean, \(\mu\) is the population mean (30 mpg), \(\sigma\) is the sample standard deviation, and \(n\) is the total number of samples (6 in this case).
03

Interpret the T-Test Results

Finally, compare the calculated t-value with the critical t-value for a one-tailed test at a certain significance level (e.g., 5%), with degrees of freedom equal to \(n - 1\). If the calculated t-value is less than the critical t-value, do not reject the null hypothesis. This means the data do not provide enough evidence to contradict the manufacturer's claim. If the calculated t-value is greater than the critical t-value, reject the null hypothesis. This means the data indicates that the true average fuel efficiency may be less than 30 mpg.

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Key Concepts

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

Sample Mean Calculation
Understanding the sample mean is crucial in statistics, as it gives us a measure of the central tendency of a data set. To compute the sample mean, add up all the data values and divide by the number of data points. For our specific fuel efficiency test example, we have six drivers with respective fuel efficiencies: 27.2, 29.3, 31.2, 28.4, 30.3, and 29.6 miles per gallon. Adding these numbers together gives us a total of 175. The sample size, denoted by n, is 6. Thus, calculating the sample mean involves the simple formula:
\[\begin{equation} \bar{x} = \frac{\text{Sum of observed values}}{n} \end{equation}\]
, which for our data set turns out to be approximately 29.17 mpg. This calculated mean is the average fuel efficiency obtained from our selected sample and will be fundamental in determining whether the automobile's claim holds up under statistical scrutiny.
One-Sample T-test
When comparing the average of a sample to a known value, the one-sample t-test serves as a robust tool. The test determines if there's a statistically significant difference between the sample mean and the established population mean. The mentioned automobile manufacturer claimed an average fuel efficiency of 30 mpg, which is our population mean, \mu. To perform this test, calculate the t-value using the formula:
\[\begin{equation} t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \end{equation}\]
, where \bar{x} is the sample mean, \mu is the population mean, s is the sample standard deviation, and n is the sample size. After calculating the t-value, we compare it to a critical value from the t-distribution table, based on our degrees of freedom (n-1) and desired level of significance, often 0.05 for a 95% confidence level. If our t-value exceeds the critical value, we reject the null hypothesis, suggesting the true mean is different from the claimed mean. By contrast, if the t-value is less than the critical value, there isn't sufficient evidence to conclude a difference, and we do not reject the null hypothesis.
Null Hypothesis Evaluation
In the realm of hypothesis testing, the null hypothesis, denoted H0, stands as a default statement asserting that there is no effect or no difference. In our fuel efficiency investigation, the null hypothesis posits that the true average fuel efficiency is at least 30 mpg, which is the value advertised by the manufacturer. Evaluating this null hypothesis involves analyzing the t-test results. If the results show that our sample mean is significantly lower than the population mean (with a calculated t-value greater than the critical value), we have grounds to reject H0. This rejection would indicate that the actual fuel efficiency is likely less than 30 mpg, contrary to the manufacturer's claim. Conversely, if the t-value does not reach the critical value, the difference is not significant, and the sample provides insufficient evidence to disprove the manufacturer’s claim. Thus, we would not reject the null hypothesis. Evaluating the null hypothesis is pivotal in statistical testing as it guides decision-makers on whether to support or discard assumptions based on empirical evidence.

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