Answer :
Let's address the given problem step-by-step:
a. State the appropriate null and alternative hypotheses.
In statistical hypothesis testing, we need to set up two hypotheses about a population parameter (in this case, the mean number of miles between oil changes). The null hypothesis ([tex]\(H_0\)[/tex]) is a statement of no effect or no difference, and it is assumed true until evidence indicates otherwise. The alternative hypothesis ([tex]\(H_A\)[/tex]) is what we want to test for.
Given the context, we want to see if the mean number of miles between oil changes is less than 3500 miles. Thus:
- The null hypothesis ([tex]\(H_0\)[/tex]) is: The mean number of miles is at least 3500 miles.
- The alternative hypothesis ([tex]\(H_A\)[/tex]) is: The mean number of miles is less than 3500 miles.
In symbolic form:
- [tex]\(H_0: \mu \geq 3500\)[/tex]
- [tex]\(H_A: \mu < 3500\)[/tex]
So, the correct answer is:
C. [tex]\(H_0: \mu \geq 3500\)[/tex], [tex]\(H_A: \mu < 3500\)[/tex]
b. Use the test statistic approach to test the null hypothesis with [tex]\(\alpha = 0.025\)[/tex].
1. Determine the critical value for [tex]\(\alpha = 0.025\)[/tex].
For a one-tailed test with [tex]\(\alpha = 0.025\)[/tex] and degrees of freedom ([tex]\(df\)[/tex]) equal to [tex]\(n - 1\)[/tex], where [tex]\(n\)[/tex] is the sample size.
Given, the sample size [tex]\(n = 10\)[/tex]:
[tex]\[ df = n - 1 = 10 - 1 = 9 \][/tex]
By looking up a t-distribution table or using statistical software for [tex]\(\alpha = 0.025\)[/tex] and [tex]\(df = 9\)[/tex], we get the critical value:
[tex]\[ \text{Critical value} = -2.2622 \][/tex]
2. Calculate the test statistic.
Let's break down the calculation of the test statistic:
- Sample mean ([tex]\(\bar{x}\)[/tex]): 3505.0
- Population mean under null hypothesis ([tex]\(\mu\)[/tex]): 3500
- Sample standard deviation ([tex]\(s\)[/tex]): 253.2566
- Sample size ([tex]\(n\)[/tex]): 10
Formula for the test statistic ([tex]\(t\)[/tex]):
[tex]\[ t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \][/tex]
Substituting the values:
[tex]\[ t = \frac{3505.0 - 3500}{253.2566 / \sqrt{10}} \][/tex]
Evaluating this expression:
[tex]\[ t = \frac{5.0}{253.2566 / 3.1623} \][/tex]
[tex]\[ t \approx \frac{5.0}{80.0915} \][/tex]
[tex]\[ t \approx 0.0624 \][/tex]
So, the test statistic is 0.0624, which rounds to:
[tex]\[ \boxed{0.06} \quad \text{(rounded to two decimal places)} \][/tex]
3. Compare the test statistic with the critical value:
- If the test statistic [tex]\((t)\)[/tex] is less than the critical value, we reject the null hypothesis.
- Here, [tex]\(t = 0.06\)[/tex], and the critical value is [tex]\(-2.2622\)[/tex].
Since [tex]\(0.06\)[/tex] is greater than [tex]\(-2.2622\)[/tex], we do not reject the null hypothesis.
Conclusion:
Given the test statistic of [tex]\(0.06\)[/tex] does not fall in the critical region, we do not have sufficient evidence to conclude that the mean number of miles between oil changes for franchise customers is less than 3500 miles. Therefore, we fail to reject the null hypothesis.
a. State the appropriate null and alternative hypotheses.
In statistical hypothesis testing, we need to set up two hypotheses about a population parameter (in this case, the mean number of miles between oil changes). The null hypothesis ([tex]\(H_0\)[/tex]) is a statement of no effect or no difference, and it is assumed true until evidence indicates otherwise. The alternative hypothesis ([tex]\(H_A\)[/tex]) is what we want to test for.
Given the context, we want to see if the mean number of miles between oil changes is less than 3500 miles. Thus:
- The null hypothesis ([tex]\(H_0\)[/tex]) is: The mean number of miles is at least 3500 miles.
- The alternative hypothesis ([tex]\(H_A\)[/tex]) is: The mean number of miles is less than 3500 miles.
In symbolic form:
- [tex]\(H_0: \mu \geq 3500\)[/tex]
- [tex]\(H_A: \mu < 3500\)[/tex]
So, the correct answer is:
C. [tex]\(H_0: \mu \geq 3500\)[/tex], [tex]\(H_A: \mu < 3500\)[/tex]
b. Use the test statistic approach to test the null hypothesis with [tex]\(\alpha = 0.025\)[/tex].
1. Determine the critical value for [tex]\(\alpha = 0.025\)[/tex].
For a one-tailed test with [tex]\(\alpha = 0.025\)[/tex] and degrees of freedom ([tex]\(df\)[/tex]) equal to [tex]\(n - 1\)[/tex], where [tex]\(n\)[/tex] is the sample size.
Given, the sample size [tex]\(n = 10\)[/tex]:
[tex]\[ df = n - 1 = 10 - 1 = 9 \][/tex]
By looking up a t-distribution table or using statistical software for [tex]\(\alpha = 0.025\)[/tex] and [tex]\(df = 9\)[/tex], we get the critical value:
[tex]\[ \text{Critical value} = -2.2622 \][/tex]
2. Calculate the test statistic.
Let's break down the calculation of the test statistic:
- Sample mean ([tex]\(\bar{x}\)[/tex]): 3505.0
- Population mean under null hypothesis ([tex]\(\mu\)[/tex]): 3500
- Sample standard deviation ([tex]\(s\)[/tex]): 253.2566
- Sample size ([tex]\(n\)[/tex]): 10
Formula for the test statistic ([tex]\(t\)[/tex]):
[tex]\[ t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \][/tex]
Substituting the values:
[tex]\[ t = \frac{3505.0 - 3500}{253.2566 / \sqrt{10}} \][/tex]
Evaluating this expression:
[tex]\[ t = \frac{5.0}{253.2566 / 3.1623} \][/tex]
[tex]\[ t \approx \frac{5.0}{80.0915} \][/tex]
[tex]\[ t \approx 0.0624 \][/tex]
So, the test statistic is 0.0624, which rounds to:
[tex]\[ \boxed{0.06} \quad \text{(rounded to two decimal places)} \][/tex]
3. Compare the test statistic with the critical value:
- If the test statistic [tex]\((t)\)[/tex] is less than the critical value, we reject the null hypothesis.
- Here, [tex]\(t = 0.06\)[/tex], and the critical value is [tex]\(-2.2622\)[/tex].
Since [tex]\(0.06\)[/tex] is greater than [tex]\(-2.2622\)[/tex], we do not reject the null hypothesis.
Conclusion:
Given the test statistic of [tex]\(0.06\)[/tex] does not fall in the critical region, we do not have sufficient evidence to conclude that the mean number of miles between oil changes for franchise customers is less than 3500 miles. Therefore, we fail to reject the null hypothesis.
