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To solve the problem given, we need to perform a hypothesis test to determine if there is a significant difference in the stopping distances between midsize and compact automobiles at a significance level of \(\alpha = 0.10\).

### Step 1: Define the Hypotheses
- **Null Hypothesis (\(H_0\))**: The mean stopping distance for midsize cars is the same as that for compact cars. \(\mu_1 = \mu_2\)
- **Alternative Hypothesis (\(H_1\))**: The mean stopping distance for midsize cars is different from that for compact cars. \(\mu_1 \neq \mu_2\)

### Step 2: Gather the Data
- **Midsize Cars Stopping Distances**: 188, 190, 195, 192, 186, 194, 188, 187, 214, 203
- **Compact Cars Stopping Distances**: 200, 211, 206, 297, 198, 204, 218, 212, 196, 193

### Step 3: Calculate the Sample Means and Standard Deviations

\[
\bar{X}_1 = \text{mean of midsize stopping distances}
\]
\[
\bar{X}_2 = \text{mean of compact stopping distances}
\]

\[
s_1 = \text{standard deviation of midsize stopping distances}
\]
\[
s_2 = \text{standard deviation of compact stopping distances}
\]

### Step 4: Conduct a Two-Sample t-test

Since the sample sizes are small (n < 30), we'll use a two-sample t-test. The test statistic is given by:

\[
t = \frac{\bar{X}_1 - \bar{X}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}
\]

Where \(n_1 = n_2 = 10\) are the sample sizes.

### Step 5: Find the Critical t-value and Compare

We'll compare the calculated t-value against the critical t-value for a two-tailed test at \(\alpha = 0.10\) and 18 degrees of freedom (since \(df = n_1 + n_2 - 2 = 18\)).

### Step 6: Conclusion

If the calculated t-value is greater than the critical t-value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.

Let's proceed with these calculations.

I'll do the calculations now.### Calculation Results:

- **Mean Stopping Distance (Midsize)**: \(193.7\)
- **Mean Stopping Distance (Compact)**: \(213.5\)
- **Standard Deviation (Midsize)**: \(8.73\)
- **Standard Deviation (Compact)**: \(30.36\)
- **Calculated t-statistic**: \(-1.982\)
- **Critical t-value**: \( \pm 1.734\) (for \(\alpha = 0.10\) and 18 degrees of freedom)
- **p-value**: \(0.063\)

### Conclusion:
- The calculated t-statistic (\(-1.982\)) is less than the critical t-value (\(\pm 1.734\)).
- The p-value \(0.063\) is slightly above the significance level \(\alpha = 0.10\).

Since the t-statistic is not within the rejection region and the p-value is greater than \(\alpha\), we **fail to reject the null hypothesis**. 

### Interpretation:
At the 0.10 significance level, there isn't sufficient evidence to suggest that there is a significant difference in the stopping distances between midsize and compact automobiles. Therefore, it may not make a significant difference in terms of stopping distance which type of automobile you purchase.

Would you like any further details or have any questions?

Here are five related questions to consider:
1. What would be the result if a different significance level was chosen?
2. How would the results change if the sample sizes were larger?
3. How does the variance in stopping distances affect the reliability of the test?
4. Could outliers in the compact car data (like the value 297) affect the test result?
5. How does this analysis influence safety decisions when purchasing an automobile?

**Tip:** Always check for outliers and consider their potential impact on statistical tests, especially in small sample sizes.

Learn how to perform a hypothesis test to compare the stopping distances between midsize and compact automobiles. Step-by-step guide with sample data, calculated means, standard deviations, and t-statistics. Explore the significance level and conclusion of the test.

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