91Ó°ÊÓ

Operators of gasoline-fueled vehicles complain about the price of gasoline in gas stations. According to the American Petroleum Institute, the federal gas tax per gallon is constant \((18.4 c\) as of January 13,2005 ), but state and local taxes vary from \(7.5 c\) to \(32.10 c\) for \(n=18\) key metropolitan areas around the country. \(^{\star}\) The total tax per gallon for gasoline at each of these 18 locations is given next. Suppose that these measurements constitute a random sample of size 18 $$\begin{aligned} &\begin{array}{llllll} 42.89 & 53.91 & 48.55 & 47.90 & 47.73 & 46.61 \end{array}\\\ &\begin{array}{llllll} 40.45 & 39.65 & 38.65 & 37.95 & 36.80 & 35.95 \end{array}\\\ &\begin{array}{llllll} 35.09 & 35.04 & 34.95 & 33.45 & 28.99 & 27.45 \end{array} \end{aligned}$$ a. Is there sufficient evidence to claim that the average per gallon gas tax is less than \(45 \%\) ? Use the \(t\) table in the appendix to bound the \(p\) -value associated with the test. b. What is the exact \(p\) -value?: c. Construct a \(95 \%\) confidence interval for the average per gallon gas tax in the United States.

Step by step solution

01

Formulate the Null and Alternative Hypotheses

We want to test if the average per gallon gas tax is less than 45. The null hypothesis is: \( H_0: \mu = 45 \) and the alternative hypothesis is: \( H_a: \mu < 45 \). This is a one-tailed test.

02

Compute Sample Mean and Standard Deviation

Calculate the sample mean \( \bar{x} \) and sample standard deviation \( s \) from the given data. The data points are: 42.89, 53.91, 48.55, 47.90, 47.73, 46.61, 40.45, 39.65, 38.65, 37.95, 36.80, 35.95, 35.09, 35.04, 34.95, 33.45, 28.99, and 27.45.- Sample mean \( \bar{x} = \frac{\sum x_i}{n} \approx 39.67 \)- Sample standard deviation \( s = \sqrt{\frac{1}{n-1} \sum (x_i - \bar{x})^2} \approx 7.10 \)

03

Calculate the Test Statistic

The test statistic for the hypothesis test is calculated using the formula:\[ t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} \]Where \( \mu_0 = 45 \), \( \bar{x} = 39.67 \), \( s = 7.10 \), and \( n = 18 \).Substituting the values:\[ t = \frac{39.67 - 45}{7.10/\sqrt{18}} \approx -3.12 \]

04

Determine the Critical Value and p-Value

Using the \( t \) distribution table with degrees of freedom \( df = n-1 = 17 \), find the critical \( t \) value for a one-tailed test at 5% significance level. The closest \( t \) critical value at df = 17 for 5% level is around -1.74. Our calculated \( t \approx -3.12 \) is less than this, indicating a significant result.Look up the \( t \)-distribution table to find the bounds of the \( p \)-value, which is \( p < 0.005 \). The exact \( p \)-value, using a calculator, is approximately 0.003.

05

Construct the Confidence Interval

To construct a 95% confidence interval for the mean gas tax, use the formula:\[ \bar{x} \pm t_{0.025} \times \frac{s}{\sqrt{n}} \]Where \( \bar{x} = 39.67 \), \( t_{0.025} \approx 2.11 \) for \( df = 17 \), and \( s = 7.10 \).The confidence interval is:\[ 39.67 \pm 2.11 \times \frac{7.10}{\sqrt{18}} \approx (36.27, 43.07) \]

06

Conclusion

Since the \( p \)-value is less than 0.05, we reject the null hypothesis, concluding there is sufficient evidence that the average gas tax is less than 45. The 95% confidence interval further supports this, since it does not contain 45.

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Key Concepts

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

Confidence Interval

A confidence interval is a range of values that is used to estimate an unknown population parameter, such as a mean. It's a way to express how certain we are about our estimate. In the gasoline tax example, we're given a 95% confidence interval. This means that if we repeated our sampling and interval calculations a large number of times, 95% of the intervals would contain the true average gas tax.

Understanding the components of a confidence interval can be quite enlightening:

• The sample mean (\( \bar{x} \approx 39.67 \)) is our best estimate of the population mean.
• The margin of error, calculated using the t-distribution, indicates how much uncertainty is in our estimate. This incorporates the sample standard deviation (\( s \approx 7.10 \)) and the size of the sample (\( n=18 \)).
• The confidence level (95% in our case) indicates the degree of certainty we have in the interval.

In this instance, the confidence interval from the exercise is approximately (36.27, 43.07). This suggests that the researcher is 95% confident the true average gas tax is between 36.27 and 43.07 cents per gallon.

t-distribution

The t-distribution is fundamental in statistics, especially when dealing with smaller sample sizes, typically less than 30. It's used instead of the normal distribution because it provides a more generous estimate of variance, accounting for the extra uncertainty from smaller samples.

The shape of the t-distribution depends on the degrees of freedom (df), which in hypothesis testing equals the sample size minus one (\( df = n - 1 \)). In our exercise, we have \( n = 18 \), leading to \( df = 17 \).

A few key characteristics of the t-distribution:

• More spread out than the normal distribution, with thicker tails. This accounts for the increased variability and uncertainty in small samples.
• As sample size increases, the t-distribution approaches the normal distribution.
• Used to calculate the critical t-value which determines the bounds of our confidence interval and our significance test.

The exercise involved determining a critical value for a one-tailed test at a 5% significance level, using the t-distribution table, finding it around -1.74 for \( df = 17 \).

p-value

The p-value measures the probability of obtaining a test statistic as extreme as the one observed, assuming the null hypothesis is true. It helps us understand the weight of evidence against the null hypothesis.

In the context of our gasoline tax test, we set the null hypothesis: \( H_0: \mu = 45 \), positing no difference from 45 cents. Our alternative hypothesis \( H_a: \mu < 45 \) proposes that the mean is less.

To get the p-value:

• Calculate the test statistic (\( t \approx -3.12 \)).
• Look up the corresponding p-value in the t-distribution table.
• In this exercise, the exact p-value was found to be approximately 0.003.

The p-value of 0.003 is quite small, less than 0.05, confirming that the evidence against the null hypothesis is strong. This means we reject the null hypothesis, supporting the claim that the mean gas tax is indeed less than 45 cents per gallon.