Let's solve this step by step. Here, we are dealing with hypothesis testing to test the manufacturer's claim about the lifespan of tires.
### 1. Formulate Hypotheses
We need to set up our null and alternative hypotheses:
- Null Hypothesis (H0): The average lifespan of the tires is at least 50,000 miles.
[tex]\[
H_0: \mu \geq 50,000
\][/tex]
- Alternative Hypothesis (H1): The average lifespan of the tires is less than 50,000 miles.
[tex]\[
H_1: \mu < 50,000
\][/tex]
### 2. Given Data
- The sample size [tex]\( n = 22 \)[/tex]
- Sample mean [tex]\( \overline{x} = 48,500 \)[/tex] miles
- Population mean [tex]\( \mu = 50,000 \)[/tex] miles
- Population standard deviation [tex]\( \sigma = 8,000 \)[/tex] miles
- Significance level [tex]\( \alpha = 0.05 \)[/tex]
### 3. Test Statistic
We use the z-test for hypothesis testing. The test statistic [tex]\( z \)[/tex] is calculated as follows:
[tex]\[
z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}
\][/tex]
Plugging in the given data:
[tex]\[
z = \frac{48,500 - 50,000}{\frac{8,000}{\sqrt{22}}} \approx -0.879
\][/tex]
### 4. p-value
The p-value is the probability of obtaining a test statistic at least as extreme as the one observed, under the null hypothesis. For a left-tailed test, we can find the p-value using the cumulative distribution function (CDF) of the standard normal distribution.
[tex]\[
p\text{-value} = P(Z < -0.879) \approx 0.1896
\][/tex]
### 5. Decision Rule
Compare the p-value to the significance level [tex]\( \alpha \)[/tex]:
- If [tex]\( p\text{-value} \leq \alpha \)[/tex], reject the null hypothesis.
- If [tex]\( p\text{-value} > \alpha \)[/tex], fail to reject the null hypothesis.
Given that our p-value [tex]\( \approx 0.1896 \)[/tex] is greater than [tex]\( \alpha = 0.05 \)[/tex]:
[tex]\[
p\text{-value} = 0.1896 > \alpha = 0.05
\][/tex]
Thus, we fail to reject the null hypothesis.
### 6. Conclusion
Based on the decision:
"There is not enough evidence to conclude that the tires last, on average, fewer miles than claimed."
This means the sample data does not provide sufficient evidence to dispute the manufacturer's claim that the tires average at least 50,000 miles before needing replacement.
