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An automobile manufacturer decides to carry out a fuel efficiency test to determine if it can advertise that one of its models achieves \(30 \mathrm{mpg}\) (miles per gallon). Six people each drive a car from Phoenix to Los Angeles. The resulting fuel efficiencies (in miles per gallon) are: \(\begin{array}{ll}30.3 & 29.6\end{array}\) \(\begin{array}{llll}27.2 & 29.3 & 31.2 & 28.4\end{array}\) Assuming that fuel efficiency is normally distributed, do these data provide evidence against the claim that actual mean fuel efficiency for this model is (at least) \(30 \mathrm{mpg}\) ?

Short Answer

Expert verified
In this problem, we perform a one-tailed t-test to determine if there's enough evidence to reject the claim that the actual mean fuel efficiency is at least 30 mpg. The null hypothesis (\(H_0\)) is \(\mu \geq 30\), and the alternative hypothesis (\(H_1\)) is \(\mu < 30\). We calculate the sample mean (\(\bar{x}\)) and standard deviation (s) for the six given data points, then find the t-score and the critical t-value for a significance level of 0.05 with five degrees of freedom (n-1). If the t-score is smaller than the critical t-value, we reject the null hypothesis. Based on the result, we conclude whether there is enough evidence to reject the claim that the actual mean fuel efficiency is at least 30 mpg.

Step by step solution

01

State the null and alternative hypotheses

The null hypothesis (\(H_0\)) is that the actual mean fuel efficiency is at least 30 mpg, while the alternative hypothesis (\(H_1\)) is that the mean fuel efficiency is less than 30 mpg. \(H_0: \mu \geq 30 mpg\) \(H_1: \mu < 30 mpg\)
02

Calculate the sample mean and standard deviation

We are given the fuel efficiencies of six cars: \(30.3, 29.6, 27.2, 29.3, 31.2,\) and \(28.4\). To calculate the sample mean (\(\bar{x}\)) and standard deviation (s), we use the following formulas: \(\bar{x} = \frac{\sum{x}}{n}\) \(s = \sqrt{\frac{\sum(x - \bar{x})^2}{n-1}}\) Let's find the sample mean and standard deviation.
03

Calculate the t-score

To find the t-score, we use the formula: \(t = \frac{\bar{x} - \mu}{s / \sqrt{n}}\) Where \(s\) is the sample standard deviation, \(\bar{x}\) is the sample mean and \(n\) is the number of sample data points. Calculate the t-score using the sample mean and standard deviation found in step 2.
04

Determine the critical t-value

Since we are performing a one-tailed test with five degrees of freedom (n-1), we can find the critical t-value at a significance level of \(\alpha\), for example, 0.05, using a t-distribution table or using software like R or Python.
05

Compare t-score with critical t-value and draw conclusions

Compare the calculated t-score with the critical t-value. If the t-score is smaller than the critical t-value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis. Based on the comparison, we can draw conclusions about whether there is sufficient evidence to reject the claim that the actual mean fuel efficiency is at least 30 mpg.

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Key Concepts

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

Null Hypothesis
The null hypothesis, often represented as \( H_0 \), is a statement used in statistics that suggests there is no effect or no change present. When performing hypothesis testing, the null hypothesis is what researchers aim to test against and possibly reject. In the context of the fuel efficiency study, the null hypothesis posits that the actual mean fuel efficiency \( \mu \) for the automobile model is at least 30 miles per gallon (mpg). This forms a benchmark for comparison, implying that the car's performance is as advertised or better.

In hypothesis testing, we start with the assumption that the null hypothesis is true. Only when we have sufficient evidence, typically through statistical calculations, can we reject the null hypothesis in favor of the alternative hypothesis.
Alternative Hypothesis
Opposite to the null hypothesis is the alternative hypothesis, denoted as \( H_1 \) or \( H_a \). It represents a statement that indicates a new effect or change, which the researcher wants to prove exists. In our fuel efficiency example, the alternative hypothesis suggests that the mean fuel efficiency is less than 30 mpg. This is what the automobile manufacturer is attempting to disprove; they hope that the evidence will not support this hypothesis and therefore retain their claim of at least 30 mpg efficiency.

This hypothesis drives the research forward. If statistical analysis provides substantial support for the alternative hypothesis, it can lead to rejecting the null hypothesis. But remember, failing to reject the null does not confirm its truth but merely indicates insufficient evidence to support the alternative claim.
T-Score Calculation
A t-score is a type of statistic calculated from a sample's mean, standard deviation, and size. It helps to determine how far away the sample mean is from the population mean in units of the standard error. You can calculate the t-score using the formula:
\[ t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \]
where \( \bar{x} \) is the sample mean, \( \mu \) is the population mean hypothesized in the null hypothesis, \( s \) is the sample standard deviation, and \( n \) is the number of observations in the sample. In our fuel efficiency case, the t-score will show how the sample data compares to the presumed population mean of 30 mpg. A t-score far away from zero may indicate that the sample mean is significantly different from the population mean.
Critical Value
The critical value in hypothesis testing is a point on the distribution that is compared with the test statistic to decide whether to reject the null hypothesis. If the test statistic is more extreme than the critical value, we reject the null hypothesis. Critical values are determined based on the desired significance level (alpha, often set at 0.05 for a 5% significance level) and the degrees of freedom in the test.

In the example given, using a t-distribution table or statistical software, we would find the critical value for a one-tailed test with five degrees of freedom. This critical value represents the boundary for which 95% of the data would fall under if the null hypothesis were true.
Sample Mean and Standard Deviation
The sample mean is the average of all the observations in a sample and is denoted by \( \bar{x} \). It serves as an unbiased estimator of the population mean. To calculate the sample mean, add all the observations together and divide by the number of observations.

Standard deviation, represented by \( s \), measures how wide the distribution of the sample data is; essentially, it's the average distance from the mean for each data point. It's crucial for calculating the standard error, which adjusts the standard deviation for sample size, establishing a consistent measure across different sample sizes.

  • Sample Mean \( \bar{x} \) = \( \frac{\sum{x}}{n} \)
  • Sample Standard Deviation \( s \) = \( \sqrt{\frac{\sum{(x - \bar{x})^2}}{n-1}} \)

These two statistics are used to determine the t-score, which informs the researchers whether or not to reject the null hypothesis in the context of our automobile manufacturer's study.

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Most popular questions from this chapter

The paper "Patterns and Composition of Weight Change in College Freshmen" (College Student Journal [2015]: \(553-564)\) reported that the freshman year weight gain for the students in a representative sample of 103 freshmen at a midwestern university was 5.7 pounds and that the standard deviation of the weight gain was 6.8 pounds. The authors also reported that \(75.7 \%\) of these students gained more than 1.1 pounds, \(17.4 \%\) maintained their weight within 1.1 pounds, and \(6.8 \%\) lost more than 1.1 pounds. a. Based on the given information, explain why it is not reasonable to think that the distribution of weight gains for the population of freshmen students at this university is approximately normal. b. Explain why it is reasonable to use a one-sample \(t\) confidence interval to estimate the population mean even though the population distribution is not approximately normal. c. Calculate and interpret a confidence interval for the population mean weight gain of freshmen students at this university.

The authors of the paper "Changesin Quantity, Spending, and Nutritional Characteristics of Adult, Adolescent and Child Urban Corner Store Purchases After an Environmental Intervention"(Preventative Medicine [2015]: \(81-85\) ) wondered if increasing the availability of healthy food options would also increase the amount people spend at the corner store. They collected data from a representative sample of 5949 purchases at corner stores in Philadelphia after the stores increased their healthy food options. The sample mean amount spent for this sample of purchases was \(\$ 2.86\) and the sample standard deviation was \(\$ 5.40\). a. Notice that for this sample, the sample standard deviation is greater than the sample mean. What does this tell you about the distribution of purchase amounts? b. Before the stores increased the availability of healthy foods, the population mean total amount spent per purchase was thought to be about \(\$ 2.80\). Do the data from this study provide convincing evidence that the population mean amount spent per purchase is greater after the change to increased healthy food options? Carry out a hypothesis test with a significance level of 0.05

USA TODAY reported that the average amount of money spent on coffee drinks each month is \(\$ 78.00\) (USA Snapshot, November 4, 2016). a. Suppose that this estimate was based on a representative sample of 20 adult Americans. Would you recommend using the one-sample \(t\) confidence interval to estimate the population mean amount spent on coffee for the population of all adult Americans? Explain why or why not. b. If the sample size had been 200 , would you recommend using the one-sample \(t\) confidence interval to estimate the population mean amount spent on coffee for the population of all adult Americans? Explain why or why not.

Give as much information as you can about the \(P\) -value of a \(t\) test in each of the following situations: a. Two-tailed test, \(n=16, t=1.6\) b. Upper-tailed test, \(n=14, t=3.2\) c. Lower-tailed test, \(n=20, t=-5.1\) d. Two-tailed test, \(n=16, t=6.3\)

The Economist collects data each year on the price of a Big Mac in various countries around the world. A sample of McDonald's restaurants in Europe in July 2016 resulted in the following Big Mac prices (after conversion to U.S. dollars): \(\begin{array}{llll}4.44 & 3.15 & 2.42 & 3.96\end{array}\) \(\begin{array}{llll}4.51 & 4.17 & 3.69 & 4.62\end{array}\) \(\begin{array}{lll}3.80 & 3.36 & 3.85\end{array}\) The mean price of a Big Mac in the U.S. in July 2016 was \$5.04. For purposes of this exercise, you can assume it is reasonable to regard the sample as representative of European McDonald's restaurants. Does the sample provide convincing evidence that the mean July 2016 price of a Big Mac in Europe is less than the reported U.S. price? Test the relevant hypotheses using \(\alpha=0.05 .\) (Hint: See Example 12.12.)

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