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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 report "Majoring in Money: How American College Students Manage Their Finances" (SallieMae, \(2016,\) www.news/salliemae.com, retrieved December 24,2106\()\) includes data from a survey of college students. Each person in a representative sample of 793 college students was asked if they had one or more credit cards and if so, whether they paid their balance in full each month. There were 500 who paid in full each month. For this sample of 500 students, the sample mean credit card balance was reported to be \(\$ 825 .\) The sample standard deviation of the credit card balances for these 500 students was not reported, but for purposes of this exercises, suppose that it was \(\$ 200\). Is there convincing evidence that college students who pay their credit card balance in full each month have mean balance that is lower than \(\$ 906\), the value reported for all college students with credit cards? Carry out a hypothesis test using a significance level of 0.01 .

Suppose that the population mean value of interpupillary distance (the distance between the pupils of the left and right eyes) for adult males is \(65 \mathrm{~mm}\) and that the population standard deviation is \(5 \mathrm{~mm}\). a. If the distribution of interpupillary distance is normal and a random sample of \(n=25\) adult males is to be selected, what is the probability that the sample mean distance \(\bar{x}\) for these 25 will be between 64 and \(67 \mathrm{~mm}\) ? At least \(68 \mathrm{~mm}\) ? b. Suppose that a random sample of 100 adult males is to be selected. Without assuming that interpupillary distance is normally distributed, what is the approximate probability that the sample mean distance will be between 64 and 67 \(\mathrm{mm}\) ? At least \(68 \mathrm{~mm} ?\)

USA TODAY (October 14,2016 ) reported that Americans spend 4.1 hours per weekday checking work e-mail. This was an estimate based on a survey of 1004 white-collar workers in the United States. a. Suppose that you would like to know if there is evidence that the mean time spent checking work e-mail for white-collar workers in the United States is more than half of the 8 -hour work day. What would you need to assume about the sample in order to use the given sample data to answer this question? b. Given that any concerns about the sample were satisfactorily addressed, carry out a test to decide if there is evidence that the mean time spent checking work e-mail for white-collar workers in the United States is more than half of the 8 -hour work day. Suppose that the sample standard deviation was \(s=1.3\) hours.

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, \(\mathrm{df}=9, t=0.73\) b. Upper-tailed test, \(\mathrm{df}=10, t=0.5\) c. Lower-tailed test, \(n=20, t=-2.1\) d. Two-tailed test, \(n=40, t=1.7\)

The formula used to calculate a confidence interval for the mean of a normal population is $$ \bar{x} \pm(t \text { critical value }) \frac{s}{\sqrt{n}} $$ What is the appropriate \(t\) critical value for each of the following confidence levels and sample sizes? a. \(95 \%\) confidence, \(n=17\) b. \(99 \%\) confidence, \(n=24\) c. \(90 \%\) confidence, \(n=13\)

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