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Sagot :
Let's consider the given problem and test whether the auto maker's claim that the transmissions can run for over 150,000 miles before failure holds true. We will use a t-test for the hypothesis testing.
### Given Data:
- Sample Mean ([tex]$\bar{x}$[/tex]): 150.7 thousand miles
- Hypothesized Mean ([tex]$\mu_0$[/tex]): 150 thousand miles
- Variance ([tex]$s^2$[/tex]): 4.551
- Sample Size ([tex]$n$[/tex]): 42
- Degrees of Freedom ([tex]$df$[/tex]): 41
- t Statistic ([tex]$t_{stat}$[/tex]): 2.17
- P(T ≤ t) one-tail: 0.018
- t Critical one-tail ([tex]$t_{critical}$[/tex] one-tail): 1.683
- P(T ≤ t) two-tail: 0.036
- t Critical two-tail ([tex]$t_{critical}$[/tex] two-tail): 2.02
- Confidence Level (95.0%): 0.665
### Hypothesis Testing:
1. Null Hypothesis ([tex]$H_0$[/tex]): [tex]$\mu \leq 150$[/tex]
This means that the average mileage before the transmission fails is 150,000 miles or less.
2. Alternative Hypothesis ([tex]$H_1$[/tex]): [tex]$\mu > 150$[/tex]
This means that the average mileage before the transmission fails is more than 150,000 miles.
### Calculation Steps:
1. Determine the t-statistic:
- The t-statistic provided is 2.17.
2. Identify the significance level:
- Common practice is to use a significance level of [tex]$\alpha = 0.05$[/tex].
3. Compare the t-statistic with the critical t-value:
- Given a one-tailed test with a significance level of 0.05 and degrees of freedom (41), the critical t-value ([tex]$t_{critical}$[/tex] one-tail) is 1.683.
4. Decision Rule:
- If the calculated t-statistic is greater than the critical t-value, we reject the null hypothesis.
### Conclusions:
- The calculated t-statistic (2.17) is greater than the critical t-value for a one-tailed test (1.683).
- The p-value for the one-tail test (0.018) is less than the significance level of 0.05.
Since the t-statistic exceeds the critical value of 1.683 and the p-value is less than 0.05, we have sufficient evidence to reject the null hypothesis. This means that the sample provides sufficient evidence to support the claim that the transmissions last more than 150,000 miles before failure.
### Given Data:
- Sample Mean ([tex]$\bar{x}$[/tex]): 150.7 thousand miles
- Hypothesized Mean ([tex]$\mu_0$[/tex]): 150 thousand miles
- Variance ([tex]$s^2$[/tex]): 4.551
- Sample Size ([tex]$n$[/tex]): 42
- Degrees of Freedom ([tex]$df$[/tex]): 41
- t Statistic ([tex]$t_{stat}$[/tex]): 2.17
- P(T ≤ t) one-tail: 0.018
- t Critical one-tail ([tex]$t_{critical}$[/tex] one-tail): 1.683
- P(T ≤ t) two-tail: 0.036
- t Critical two-tail ([tex]$t_{critical}$[/tex] two-tail): 2.02
- Confidence Level (95.0%): 0.665
### Hypothesis Testing:
1. Null Hypothesis ([tex]$H_0$[/tex]): [tex]$\mu \leq 150$[/tex]
This means that the average mileage before the transmission fails is 150,000 miles or less.
2. Alternative Hypothesis ([tex]$H_1$[/tex]): [tex]$\mu > 150$[/tex]
This means that the average mileage before the transmission fails is more than 150,000 miles.
### Calculation Steps:
1. Determine the t-statistic:
- The t-statistic provided is 2.17.
2. Identify the significance level:
- Common practice is to use a significance level of [tex]$\alpha = 0.05$[/tex].
3. Compare the t-statistic with the critical t-value:
- Given a one-tailed test with a significance level of 0.05 and degrees of freedom (41), the critical t-value ([tex]$t_{critical}$[/tex] one-tail) is 1.683.
4. Decision Rule:
- If the calculated t-statistic is greater than the critical t-value, we reject the null hypothesis.
### Conclusions:
- The calculated t-statistic (2.17) is greater than the critical t-value for a one-tailed test (1.683).
- The p-value for the one-tail test (0.018) is less than the significance level of 0.05.
Since the t-statistic exceeds the critical value of 1.683 and the p-value is less than 0.05, we have sufficient evidence to reject the null hypothesis. This means that the sample provides sufficient evidence to support the claim that the transmissions last more than 150,000 miles before failure.
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