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Answer :
To solve this question, we'll determine if the mean braking distances for Make A and Make B are statistically different using a hypothesis test for the difference between two means with known population standard deviations. We'll follow these steps:
### Step (a): State the Hypotheses
- Null Hypothesis ([tex]\(H_0\)[/tex]): The mean braking distances for Make A and Make B are equal. That is, [tex]\(\mu_1 = \mu_2\)[/tex].
- Alternative Hypothesis ([tex]\(H_a\)[/tex]): The mean braking distances are different. That is, [tex]\(\mu_1 \neq \mu_2\)[/tex].
### Step (b): Significance Level
The significance level ([tex]\(\alpha\)[/tex]) is 0.10. This is a two-tailed test because we are checking for any difference (not a specific direction).
### Step (c): Calculate the Standardized Test Statistic [tex]\(z\)[/tex]
Given data:
- Mean for Make A ([tex]\(\bar{x}_1\)[/tex]) = 42 feet
- Mean for Make B ([tex]\(\bar{x}_2\)[/tex]) = 44 feet
- Population standard deviation for Make A ([tex]\(\sigma_1\)[/tex]) = 4.5 feet
- Population standard deviation for Make B ([tex]\(\sigma_2\)[/tex]) = 4.2 feet
- Sample sizes are both 35.
To find the standardized test statistic [tex]\(z\)[/tex]:
1. Calculate the Standard Error of the Difference Between Means:
[tex]\[
SE_{diff} = \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}
\][/tex]
Plugging in the values:
[tex]\[
SE_{diff} = \sqrt{\frac{4.5^2}{35} + \frac{4.2^2}{35}} \approx 1.04
\][/tex]
2. Calculate the [tex]\(z\)[/tex]-score:
[tex]\[
z = \frac{\bar{x}_1 - \bar{x}_2}{SE_{diff}}
\][/tex]
[tex]\[
z = \frac{42 - 44}{1.04} \approx -1.92
\][/tex]
### Step (d): Determine the Rejection Region
For a two-tailed test with [tex]\(\alpha = 0.10\)[/tex], the critical [tex]\(z\)[/tex] values are found using the standard normal distribution table. The rejection regions are:
- [tex]\(z < -1.64\)[/tex]
- [tex]\(z > 1.64\)[/tex]
### Step (e): Make a Decision
Compare the calculated [tex]\(z\)[/tex]-score with the rejection regions:
- The calculated [tex]\(z\)[/tex] is approximately -1.92.
- Since -1.92 is less than -1.64, it falls in the lower rejection region.
### Conclusion
Since the [tex]\(z\)[/tex]-score falls into the rejection region, we reject the null hypothesis. This suggests that there is enough evidence to support the claim that the mean braking distances are different for Make A and Make B.
### Step (a): State the Hypotheses
- Null Hypothesis ([tex]\(H_0\)[/tex]): The mean braking distances for Make A and Make B are equal. That is, [tex]\(\mu_1 = \mu_2\)[/tex].
- Alternative Hypothesis ([tex]\(H_a\)[/tex]): The mean braking distances are different. That is, [tex]\(\mu_1 \neq \mu_2\)[/tex].
### Step (b): Significance Level
The significance level ([tex]\(\alpha\)[/tex]) is 0.10. This is a two-tailed test because we are checking for any difference (not a specific direction).
### Step (c): Calculate the Standardized Test Statistic [tex]\(z\)[/tex]
Given data:
- Mean for Make A ([tex]\(\bar{x}_1\)[/tex]) = 42 feet
- Mean for Make B ([tex]\(\bar{x}_2\)[/tex]) = 44 feet
- Population standard deviation for Make A ([tex]\(\sigma_1\)[/tex]) = 4.5 feet
- Population standard deviation for Make B ([tex]\(\sigma_2\)[/tex]) = 4.2 feet
- Sample sizes are both 35.
To find the standardized test statistic [tex]\(z\)[/tex]:
1. Calculate the Standard Error of the Difference Between Means:
[tex]\[
SE_{diff} = \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}
\][/tex]
Plugging in the values:
[tex]\[
SE_{diff} = \sqrt{\frac{4.5^2}{35} + \frac{4.2^2}{35}} \approx 1.04
\][/tex]
2. Calculate the [tex]\(z\)[/tex]-score:
[tex]\[
z = \frac{\bar{x}_1 - \bar{x}_2}{SE_{diff}}
\][/tex]
[tex]\[
z = \frac{42 - 44}{1.04} \approx -1.92
\][/tex]
### Step (d): Determine the Rejection Region
For a two-tailed test with [tex]\(\alpha = 0.10\)[/tex], the critical [tex]\(z\)[/tex] values are found using the standard normal distribution table. The rejection regions are:
- [tex]\(z < -1.64\)[/tex]
- [tex]\(z > 1.64\)[/tex]
### Step (e): Make a Decision
Compare the calculated [tex]\(z\)[/tex]-score with the rejection regions:
- The calculated [tex]\(z\)[/tex] is approximately -1.92.
- Since -1.92 is less than -1.64, it falls in the lower rejection region.
### Conclusion
Since the [tex]\(z\)[/tex]-score falls into the rejection region, we reject the null hypothesis. This suggests that there is enough evidence to support the claim that the mean braking distances are different for Make A and Make B.
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