A car has a mean efficiency of 32 miles per gallon (mpg) with a standard deviation of 10 mpg. A roof rack was recently added and I suspect this will lower the efficiency of the car. 31 days after the installation of the roof rack were randomly selected and the mean efficiency of this sample was 29 mpg with standard deviation of 8 mpg. Test my suspicions using a 96% confidence level.

To test your suspicion that the roof rack lowered the car’s fuel efficiency, we can perform a hypothesis test. Since we're dealing with a sample and we know the population standard deviation, we will use a **one-sample t-test**.

### Hypothesis Setup

- **Null Hypothesis (H₀):** The mean efficiency after adding the roof rack is the same as the original mean, i.e., no effect.  
  \( H_0: \mu = 32 \, \text{mpg} \)

- **Alternative Hypothesis (H₁):** The mean efficiency after adding the roof rack is less than the original mean, i.e., the roof rack lowers the efficiency.  
  \( H_1: \mu < 32 \, \text{mpg} \)

### Given Data

- Population mean efficiency before the roof rack: \( \mu_0 = 32 \, \text{mpg} \)
- Sample mean efficiency after the roof rack: \( \bar{x} = 29 \, \text{mpg} \)
- Sample standard deviation: \( s = 8 \, \text{mpg} \)
- Sample size: \( n = 31 \)
- Confidence level: 96% → **significance level** \( \alpha = 1 - 0.96 = 0.04 \)
- Degrees of freedom for the t-test: \( df = n - 1 = 31 - 1 = 30 \)

### Test Statistic (t)

The formula for the t-statistic is:

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

Substitute the values:

\[
t = \frac{29 - 32}{8 / \sqrt{31}} = \frac{-3}{8 / 5.568} = \frac{-3}{1.437} \approx -2.09
\]

### Critical Value

For a one-tailed t-test with a confidence level of 96%, the critical value at \( \alpha = 0.04 \) and 30 degrees of freedom can be found using a t-distribution table or a calculator. From the t-table or software, the critical value for \( t_{0.04, 30} \) is approximately:

\[
t_{\text{critical}} \approx -1.697
\]

### Decision Rule

- If \( t \) is less than \( t_{\text{critical}} \), we reject the null hypothesis.
- If \( t \) is greater than \( t_{\text{critical}} \), we fail to reject the null hypothesis.

Since our calculated \( t \)-value of \( -2.09 \) is less than the critical value of \( -1.697 \), we **reject the null hypothesis**.

### Conclusion

There is sufficient evidence at the 96% confidence level to support the suspicion that the roof rack lowers the fuel efficiency of the car.

---

Would you like further details on how the test was performed or explanations on hypothesis testing? Here are some related questions:

1. How does a one-tailed t-test differ from a two-tailed t-test?
2. What does the p-value represent in hypothesis testing?
3. How would you interpret the results if the test statistic were close to the critical value?
4. What are the assumptions made in performing a t-test?
5. What impact would increasing the sample size have on the test?

**Tip:** When performing hypothesis tests with smaller sample sizes (less than 30), always use the t-distribution, as the sample distribution of the test statistic may not follow a normal distribution.

Learn how to perform a one-sample t-test to test whether a roof rack installation reduces the fuel efficiency of a car. With a sample size of 31, a mean of 29 mpg, and a standard deviation of 8 mpg, this problem demonstrates how to calculate the test statistic and critical value to make data-driven conclusions at a 96% confidence level.

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