Problem 13 - Basic Statistics for Business and Economics with Student

Question

A spark plug manufacturer claimed that its plugs have a mean life in excess of 22,100 miles. Assume the life of the spark plugs follows the normal distribution. A fleet owner purchased a large number of sets. A sample of 18 sets revealed that the mean life was 23,400 miles and the standard deviation was 1,500 miles. Is there enough evidence to substantiate the manufacturer's claim at the .05 significance level?

Step-by-Step Solution

Answer

Yes, there is sufficient evidence to support the claim that the mean life is greater than 22,100 miles.

Step 1: Set Up Hypotheses

Determine the null and alternative hypotheses. The null hypothesis \( H_0 \) is that the mean life of spark plugs \( \mu \) is 22,100 miles. The alternative hypothesis \( H_a \) is that the mean life is greater than 22,100 miles. Mathematically, these are written as: \( H_0: \mu = 22,100 \) and \( H_a: \mu > 22,100 \).

Step 2: Determine the Test Statistic

Since the sample size is 18, we use a t-test for the mean. The test statistic \( t \) is calculated using the formula: \[ t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \] where \( \bar{x} = 23,400 \), \( \mu = 22,100 \), \( s = 1,500 \), and \( n = 18 \). Substituting the values, we get: \( t = \frac{23,400 - 22,100}{1,500 / \sqrt{18}} \).

Step 3: Calculate the Test Statistic

Perform the calculation from Step 2: \( t = \frac{1,300}{1,500 / \sqrt{18}} = \frac{1,300}{353.55} \approx 3.676 \).

Step 4: Determine the Critical Value

For a one-tailed t-test at \( \alpha = 0.05 \) and 17 degrees of freedom (since \( n-1 = 17 \)), we look up the critical t value from the t-distribution table. The critical value is approximately 1.740.

Step 5: Make a Decision

Compare the calculated t-statistic to the critical value. Since 3.676 > 1.740, we reject the null hypothesis.

Step 6: State the Conclusion

Since we reject the null hypothesis, there is sufficient evidence at the 0.05 significance level to support the claim that the mean life of the spark plugs is greater than 22,100 miles.

Key Concepts

t-test

The t-test is a statistical tool we use to determine if there is a significant difference between the means of two groups. It's especially useful when dealing with small sample sizes, typically under 30. In this context, we're focusing on a one-sample t-test, which is utilized to compare the sample mean to a known value or population mean.

In the exercise, we have a sample of 18 spark plug sets, with a mean life of 23,400 miles, and we use a t-test to evaluate the manufacturer's claim that their products have a mean life exceeding 22,100 miles. The test statistic is calculated using the formula:

• \[ t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \]

where \( \bar{x} \) is the sample mean, \( \mu \) is the claimed population mean, \( s \) is the sample standard deviation, and \( n \) is the sample size. By plugging these values into the formula, we determine the t-score, which will help us in deciding whether or not to reject our null hypothesis.

Significance Level

The significance level, often denoted by \( \alpha \), is a threshold we set to determine whether we can reject the null hypothesis. It represents the probability of rejecting the null hypothesis when it is actually true, which is known as a Type I error. Common levels of significance are 0.05, 0.01, and 0.10.

In our exercise, a significance level of 0.05 is used, meaning there's a 5% risk of concluding that the mean life of the spark plugs is greater than 22,100 miles when it might not be so. This level helps in assessing the evidence provided by the t-statistic calculation against the critical value. If the computed t-value exceeds the critical value determined by the significance level, we conclude there is strong enough evidence to reject the null hypothesis.

Critical Value

Critical values are threshold numbers that define the boundary between accepting and rejecting the null hypothesis. They are determined based on the significance level and the degrees of freedom, which in this case, is the sample size minus one (\( n - 1 \)). Using a t-distribution table, we find the critical value corresponding to our chosen \( \alpha \) level and 17 degrees of freedom.

For our spark plug exercise, the critical value at a 0.05 significance level for a one-tailed test is approximately 1.740. This number is vital for our analysis, as it tells us the minimum t-score needed to reject the null hypothesis. Since our computed t-score of 3.676 is greater than the critical value, it leads us to reject the null hypothesis and support the manufacturer's claim.

Normal Distribution

Normal distribution is a common assumption in statistical tests, and it describes how the probability of data points is distributed across a range of outcomes. It is often visualized as a bell-shaped curve, where most data points cluster around the mean, and the likelihood of extreme values diminishes as you move further from the center.

In the given exercise, we assumed that the life of spark plugs follows a normal distribution. This assumption allows us to use a t-test to make inferences about the population mean based on our sample mean.

• The central limit theorem supports this assumption when dealing with large samples, but even with smaller samples like 18, as in this case, the t-test can still be reliable if the data isn't severely skewed.

Understanding normal distribution is crucial as it lays the groundwork for making statistical inferences and interpreting data behaviors in real-world scenarios.