To calculate the p-value, we will use the t-test since we have a sample standard deviation and not the population standard deviation. Here’s the step-by-step process:

Given Information

• Null Hypothesis (H0): $\mu = 50$
• Alternative Hypothesis (H1): $\mu \neq 50$
• Sample mean ($\bar{x}$) = 51
• Sample size ($n$) = 30
• Sample standard deviation ($s$) = 12
• Significance level ($\alpha$) = 0.05

Step 1: Calculate the Test Statistic (t)

The formula for the test statistic $t$ in this case is: $t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}$

Plugging in the values: $t = \frac{51 - 50}{\frac{12}{\sqrt{30}}}$

First, calculate the standard error ($\frac{s}{\sqrt{n}}$): $\frac{12}{\sqrt{30}} \approx 2.1909$

Then calculate $t$: $t = \frac{1}{2.1909} \approx 0.4564$

Step 2: Determine Degrees of Freedom

The degrees of freedom ($df$) for this test is $n - 1$, so: $df = 30 - 1 = 29$

Step 3: Find the p-value

With a two-tailed test at $t = 0.4564$ and $df = 29$, we can look up the p-value using a t-distribution table or calculator. Using a t-distribution calculator, we find:

$p \text{-value} \approx 0.6514$

Answer

The p-value for this test is approximately 0.6514 (rounded to four decimal places).

Since the p-value (0.6514) is greater than the significance level (0.05), we would not reject the null hypothesis. There is not enough evidence to suggest that the average monthly gasoline expenditure differs from $50.

Would you like further details or have any questions?

Here are five related questions to explore:

1. How would the p-value change if the sample size were larger?
2. What if the sample mean was higher than 51? How would that affect the result?
3. How would using a 1% significance level change our conclusion?
4. Why is a two-tailed test appropriate in this scenario?
5. What is the difference between a one-tailed and a two-tailed hypothesis test?

Tip: Always check if the test is one-tailed or two-tailed, as it directly affects the interpretation of the p-value.