Problem 33 - Basic Statistics for Business and Economics with Student

Question

NBC TV news, in a segment on the price of gasoline, reported last evening that the mean price nationwide is 2.50 per gallon for self-serve regular unleaded. A random sample of 35 stations in the Milwaukee, Wisconsin, area revealed that the mean price was \(\$ 2.52\) per gallon and that the standard deviation was 0.05 per gallon. At the .05 significance level, can we conclude that the price of gasoline is higher in the Milwaukee area? Determine the p -value.

Step-by-Step Solution

Answer

The Milwaukee gas price is significantly higher than the national average.

Step 1: Define Hypotheses

We start by defining the null hypothesis and the alternative hypothesis. The null hypothesis \( H_0 \) states that the mean price in Milwaukee is equal to the national mean, \( \mu = 2.50 \). The alternative hypothesis \( H_a \) claims that the mean price in Milwaukee is greater, \( \mu > 2.50 \).

Step 2: Determine Test Statistic Formula

Since the population standard deviation is unknown and the sample size is 35 (greater than 30), we use a t-test. The test statistic formula is \( t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \), where \( \bar{x} \) is the sample mean, \( \mu \) is the population mean, \( s \) is the sample standard deviation, and \( n \) is the sample size.

Step 3: Calculate Test Statistic

Plugging in the values, we calculate: \( t = \frac{2.52 - 2.50}{0.05/\sqrt{35}} \). Simplify the denominator first: \( 0.05 / \sqrt{35} \approx 0.00845 \). Now, compute the entire expression: \( t \approx \frac{0.02}{0.00845} \). Thus, \( t \approx 2.37 \).

Step 4: Determine Critical Value and p-value

We use a t-distribution table to find the critical t-value. For \( n - 1 = 34 \) degrees of freedom at \( \alpha = 0.05 \) for a one-tail test, the critical value is approximately 1.691. Finding the p-value requires using a t-distribution chart or calculator, resulting in \( p \approx 0.012 \).

Step 5: Compare p-value to Significance Level

Finally, compare the p-value to the significance level \( \alpha = 0.05 \). Since \( p = 0.012 < 0.05 \), we reject the null hypothesis. This implies that there is significant evidence to conclude the mean price of gasoline is higher in Milwaukee than the national average.

Key Concepts

Null Hypothesis

In hypothesis testing, the null hypothesis is a starting point that assumes no effect or no difference exists. When we talk about testing hypotheses, the null hypothesis, denoted as \( H_0 \), is the claim we seek evidence against. In our exercise, the null hypothesis claims that the mean price of gasoline in Milwaukee is the same as the national average, which is \( \mu = 2.50 \).
This means we are initially assuming there's no difference in gasoline prices. The alternative hypothesis, denoted as \( H_a \), takes the opposite stance. It suggests that the mean price in Milwaukee is actually greater, which implies \( \mu > 2.50 \).
The purpose of conducting these tests is to see if the data we collect leads us to reject the null hypothesis. If the evidence is strong enough against it, we conclude there is some effect or difference, supporting the alternative hypothesis. Learning to formulate and understand \( H_0 \) is vital, as it shapes how we interpret data and results.

T-test

A t-test is a type of statistical test used when we want to compare a sample mean to a known population mean to determine if there is a significant difference between them. It's particularly useful when the population standard deviation is unknown and the sample size is small, though in our case, the sample size is 35, which is large enough to apply a t-test.
In our exercise, we use the formula:

• \( t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \)
• where \( \bar{x} \) is the sample mean, \( \mu \) is the population mean, \( s \) is the sample standard deviation, and \( n \) is the sample size.

This formula helps us determine how far the sample mean is from the population mean in terms of standard error units. Essentially, the t-test calculates if the observed difference is statistically significant or possibly due to random chance. In our case, after calculating, the t-value was approximately 2.37. This value tells us where the sample mean is on a standard t-distribution. If it falls in the extreme end of the distribution, we have evidence against the null hypothesis.

P-value Calculation

P-value is a key component in hypothesis testing. It tells us the probability of obtaining a result at least as extreme as the observed one, under the assumption that the null hypothesis is true. A smaller p-value indicates that the observed data would be unlikely if the null hypothesis were true, thus providing evidence against the null hypothesis.
For the gasoline price exercise, after calculating the t-score, we determine the p-value using a t-distribution chart or a calculator. With 34 degrees of freedom (since our sample size \( n = 35 \)), we found a p-value of approximately 0.012.

• If the p-value is less than the significance level \( \alpha = 0.05 \), we reject the null hypothesis.
• In our scenario, since \( p = 0.012 \) which is less than 0.05, we have strong evidence to conclude that the mean gasoline price in Milwaukee is indeed higher than the national average.

Understanding the p-value is crucial as it helps us make decisions supported by statistical evidence, rather than relying solely on observed differences. This conclusion guides researchers and decision-makers by indicating when an effect is significant enough to warrant a change in understanding or policy.