Question

A simple random sample of 40 recorded speeds (in mi/h) is obtained from cars traveling on a specific highway. The sample has a mean of 68.4 mi/h and a standard deviation of 5.7 mi/h. Use a 0.05 significance level to test the claim that the mean speed of all cars is greater than the posted speed limit of 65 mi/h. State the initial and final conclusion.

A) Reject the null hypothesis; there is not sufficient evidence to warrant a rejection of the claim that the mean speed is greater than 65 mi/h.
B) Fail to reject the null hypothesis; there is sufficient evidence to warrant a rejection of the claim that the mean speed is less than 65 mi/h.
C) Fail to reject the null hypothesis; there is not sufficient evidence to warrant a rejection of the claim that the mean speed is less than 65 mi/h.
D) Reject the null hypothesis; there is sufficient evidence to support the claim that the mean speed is greater than 65 mi/h.

Answer 1

To test the claim that the mean speed of all cars is greater than the posted speed limit of 65 mi/h, we can conduct a one-sample t-test. The hypothesis statements are formulated as follows:

- **Null hypothesis (H0):** μ ≤ 65 (The mean speed is less than or equal to 65 mi/h)
- **Alternative hypothesis (H1):** μ > 65 (The mean speed is greater than 65 mi/h)

### Step 1: Determine the test statistic

Since we know the sample mean, sample standard deviation, and the sample size, we can calculate the t-statistic using the formula:

\[
t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}
\]

Where:
- \( \bar{x} \) is the sample mean (68.4 mi/h)
- \( \mu_0 \) is the hypothesized population mean (65 mi/h)
- \( s \) is the sample standard deviation (5.7 mi/h)
- \( n \) is the sample size (40)

Plugging in the values:

\[
t = \frac{68.4 - 65}{5.7 / \sqrt{40}}
\]

Calculate the denominator:

\[
s / \sqrt{n} = \frac{5.7}{\sqrt{40}} \approx \frac{5.7}{6.3246} \approx 0.901
\]

Now substitute this back into the t statistic formula:

\[
t \approx \frac{3.4}{0.901} \approx 3.77
\]

### Step 2: Determine the critical value

Using a significance level of 0.05 for a one-tailed test and degrees of freedom (df = n - 1 = 39), we can find the critical t-value from the t-distribution table.

Using a t-table or calculator:
- Critical t-value (one-tailed, df = 39) at α = 0.05 is approximately 1.685.

### Step 3: Make the decision

Compare the calculated t-statistic to the critical t-value:
- Calculated t = 3.77
- Critical t ≈ 1.685

Since \( 3.77 > 1.685 \), we reject the null hypothesis.

### Step 4: Conclusion

The conclusion is as follows:

- We reject the null hypothesis, which indicates that there is sufficient evidence to support the claim that the mean speed of all cars is greater than 65 mi/h.

Thus, the final conclusion corresponds to option **D**:

**D) Reject the null hypothesis; there is sufficient evidence to support the claim that the mean speed is greater than 65 mi/h.**