Problem 5 - Basic Statistics for Business and Economics with Student

Question

The manufacturer of the X-15 steel-belted radial truck tire claims that the mean mileage the tire can be driven before the tread wears out is 60,000 miles. The population standard deviation of the mileage is 5,000 miles. Crosset Truck Company bought 48 tires and found that the mean mileage for its trucks is 59,500 miles. Is Crosset's experience different from that claimed by the manufacturer at the .05 significance level?

Step-by-Step Solution

Answer

Crosset's experience is not significantly different from the manufacturer's claim.

Step 1: State the Hypotheses

We're conducting a hypothesis test to determine if Crosset Truck Company's observed mean mileage differs from the manufacturer's claimed mean. The null hypothesis is that the mean mileage is 60,000 miles: \( H_0: \mu = 60,000 \). The alternative hypothesis is that the mean mileage is different from 60,000 miles: \( H_a: \mu eq 60,000 \).

Step 2: Determine the Significance Level and Test Type

The significance level is given as \( \alpha = 0.05 \). Since we're testing if the mileage is different from 60,000 miles, we use a two-tailed test.

Step 3: Calculate the Test Statistic

The test statistic for a known population standard deviation is calculated using the formula: \[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \] where \( \bar{x} = 59,500 \), \( \mu = 60,000 \), \( \sigma = 5,000 \), and \( n = 48 \). Substituting in the values, we get: \[ z = \frac{59,500 - 60,000}{\frac{5,000}{\sqrt{48}}} \approx -0.693 \].

Step 4: Find the Critical Value

For a two-tailed test at a significance level of \( \alpha = 0.05 \), the critical z-values are \( \pm 1.96 \). These values define the rejection regions in the standard normal distribution.

Step 5: Make a Decision

The calculated z-value is approximately \(-0.693\), which falls between the critical values \(-1.96\) and \(1.96\). Since it does not fall in the rejection region, we fail to reject the null hypothesis.

Step 6: Conclusion

There is not enough statistical evidence at the 0.05 significance level to conclude that Crosset Truck Company's mean mileage is different from the manufacturer's claimed mean of 60,000 miles.

Key Concepts

Null Hypothesis

In hypothesis testing, the null hypothesis is a statement that suggests there is no effect or no difference in the context of a statistical test. It is a kind of default or starting assumption. For example, in this truck tire exercise, the manufacturer's claim is that the mean mileage is 60,000 miles. Thus, our null hypothesis (H_0) is:

• \( H_0: \mu = 60,000 \text{ miles} \)

This assumption is what we are testing against, attempting to either prove wrong or show insignificance. If our statistical test does not provide enough evidence to reject the null hypothesis, we continue to assume it is true. This means that Crosset Truck Company cannot definitively say their tire mileage differs from 60,000 miles if the null hypothesis holds.

Alternative Hypothesis

While the null hypothesis posits no change or effect, the alternative hypothesis proposes that there is indeed a difference. It represents the outcome that researchers are interested in proving. For the tire mileage problem, we are checking if Crosset's mean mileage is different from the claimed figure. Thus, our alternative hypothesis (H_a) is:

• \( H_a: \mu eq 60,000 \text{ miles} \)

This is a two-tailed test because we're interested in any significant difference, either higher or lower, from the given mean of 60,000 miles. If the statistical test supports this hypothesis, it would suggest that the actual mileage differs from the manufacturer's claim.

Significance Level

The significance level, denoted by \( \alpha \), is a predetermined threshold that defines how strong the evidence must be before we reject the null hypothesis. Expressed as a decimal, it reflects the probability of wrongly rejecting the null hypothesis if it is true. In this case, the significance level is set at 0.05:

• This means we allow a 5% risk of concluding that Crosset's mean mileage is different when, in fact, it is not.
• A common significance level, 0.05, is often chosen for its balance between being too lenient or too strict.
• This level determines the critical values that will mark the rejection regions on the standard normal distribution (z-distribution).

In our exercise, the critical values for a two-tailed test with \(\alpha = 0.05\) are approximately \(\pm 1.96\). This means any test statistic beyond these values would lead us to reject the null hypothesis.

Z-Test

The z-test is a statistical test used to determine if there is a significant difference between sample data and the population mean. It is particularly applicable when the population standard deviation is known and the sample size is large. For the truck tire case, the z-test is deployed because we have:- A known population standard deviation of mileage,- A sample size that is considered large enough (48 tires).The formula for calculating the z-test statistic is given as:\[z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}\]Here:

• \(\bar{x} = 59,500\) is the sample mean,
• \(\mu = 60,000\) is the population mean,
• \(\sigma = 5,000\) is the population standard deviation,
• \(n = 48\) is the sample size.

Substituting these into the formula gives us \(z \approx -0.693\). Since this calculated z-value falls between \(\pm 1.96\), we do not reject the null hypothesis. This suggests Crosset's experience is not significantly different from the manufacturer's claim based on this data.