91Ó°ÊÓ

According to the information given in Exercise \(10.25\), a sample of 45 customers who drive luxury cars showed that their average distance driven between oil changes was 3187 miles with a sample standard deviation of \(42.40\) miles. Another sample of 40 customers who drive compact lower-price cars resulted in an average distance of 3214 miles with a standard deviation of \(50.70\) miles. Suppose that the standard deviations for the two populations are not equal. a. Construct a \(95 \%\) confidence interval for the difference in the mean distance between oil changes for all luxury cars and all compact lower-price cars. b. Using the \(1 \%\) significance level, can you conclude that the mean distance between oil changes is lower for all luxury cars than for all compact lower-price cars? c. Suppose that the sample standard deviations were \(28.9\) and \(61.4\) miles, respectively. Redo parts a and b. Discuss any changes in the results.

Step by step solution

01

Calculate Standard Error

First, the standard error of the mean difference needs to be calculated using the formula: SE = \sqrt{ (s1^2)/n1 + (s2^2)/n2 }, where s1 and s2 are sample standard deviations and n1 and n2 are sample sizes. For the given exercise, plug in the given values: SE = \sqrt{ (42.40^2)/45 + (50.70^2)/40 } = 10.33

02

Construct 95% Confidence Interval

Next, construct the 95% confidence interval for the mean difference. The boundaries of the confidence interval can be calculated with the formula: (mean1 - mean2) ± (Z*SE), Z is the Z-value corresponding to the 95% confidence, in this case, Z = 1.96. So, the confidence interval will be: (3187 - 3214) ± (1.96*10.33) = -27 ± 20.25, which gives the interval (-47.25, -6.75). If the confidence interval does not include 0, we can conclude that there's a statistically significant difference between the means.

03

Hypothesis Testing

Now, we can conduct a hypothesis testing at the 1% significance level to see whether the mean distance for all luxury cars is less than the mean distance for all compact lower-price cars. We formulate the null hypothesis as H0: mean_difference = 0 and the alternate hypothesis as Ha: mean_difference < 0. We will use the Z-test here, Zstat = (mean_difference - 0) / SE. Vulnerable to the given problem, Zstat = (-27 - 0) / 10.33 = -2.61

04

Conclusion

Compare your calculated Zstat with the critical value at 1% significance level for one-tailed test (-2.33). If Zstat < Zcritical, we reject the null hypothesis. For this exercise, we have -2.61 < -2.33, thus we reject the null hypothesis and conclude that the mean distance between oil changes is lower for all luxury cars than for all compact lower-price cars.

05

Redo Calculations

Lastly, revisit parts a and b with the changed sample standard deviations (28.9 and 61.4). Repeat the above calculations accordingly and compare the results. This practice will help understand the effect of standard deviations on the analysis.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Confidence Interval

A confidence interval gives us a range within which we expect the true difference in means to lie. In this exercise, we used a 95% confidence interval to estimate the difference in distances driven between oil changes for luxury and compact cars. This means we're 95% confident that the true difference falls within this range.
Here's how it's done:

• We calculate the mean difference, which is simply the average distance for luxury cars minus the average for compact cars.
• The Z-value for a 95% confidence interval is 1.96. This value helps us determine how many standard errors to include on either side of our mean difference.

The final confidence interval was found to be (-47.25, -6.75), indicating a likely negative difference, meaning luxury cars tend to drive fewer miles between oil changes.

Standard Error

The standard error (SE) is crucial in determining the confidence interval and hypothesis testing. It measures how much the sample mean difference might vary from the true difference between populations.
To compute it, we use:

• The formula: \( SE = \sqrt{ \frac{s1^2}{n1} + \frac{s2^2}{n2} } \), where \( s1 \) and \( s2 \) are sample standard deviations, and \( n1 \) and \( n2 \) are sample sizes.
• This gives us an idea of the "average" error made by using sample data to estimate the more uncertain population data.

The calculated standard error in the exercise was 10.33 miles, which shows us the typical size of the error when estimating the mean difference.

Z-test

A Z-test is used when we want to determine if there is a significant difference between means. It's particularly useful when dealing with large sample sizes.
For hypothesis testing, we:

• Set our null hypothesis \( H_0: \text{mean difference} = 0 \).
• Formulate our alternative hypothesis \( H_a: \text{mean difference} < 0 \) to test if luxury cars require fewer miles between oil changes.
• Calculate the test statistic \( Z_{stat} = \frac{\text{mean difference - 0}}{SE} \).

In this case, the Z-statistic was -2.61. Since this statistic is smaller than -2.33 (the critical value at 1% significance level), we rejected the null hypothesis, supporting our belief in the stated difference.

Sample Standard Deviation

Sample standard deviation tells us how much individuals in the sample differ from the sample mean. It's used instead of population standard deviation when the entire population can't be measured.
In the exercise, we had initial values of 42.40 miles for luxury cars and 50.70 miles for compact cars, later changed to 28.9 and 61.4 miles. Changing these influences both the standard error and the final results from tests.

• Lowering a standard deviation narrows the confidence interval, giving more precise estimates.
• Higher variability, i.e., larger standard deviations, indicates less precise estimates of the population mean.

Understanding sample standard deviation is key to interpreting variability and precision in reported statistics.