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The standard recommendation for automobile oil changes is once every 3000 miles. A local mechanic is interested in determining whether people who drive more expensive cars are more likely to follow the recommendation. Independent random samples of 45 customers who drive luxury cars and 40 customers who drive compact lower-price cars were selected. The average distance driven between oil changes was 3187 miles for the luxury car owners and 3214 miles for the compact lower-price cars. The sample standard deviations were \(42.40\) and \(50.70\) miles for the luxury and compact groups, respectively. Assume that the population distributions of the distances between oil changes have the same standard deviation for the two populations. a. Construct a \(95 \%\) confidence interval for the difference in the mean distances 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 less for all luxury cars than that for all compact lower-price cars?

Short Answer

Expert verified
The 95% confidence interval for the difference in mean distance between oil changes for all luxury cars and all compact lower-price cars is [-50.04, -3.96] miles, and yes, at the 1% significance level, we can conclude that the mean distance between oil changes is less for all luxury cars than that for all compact lower-price cars.

Step by step solution

01

Calculate the difference of means and standard error

First, calculate the difference of sample means which is \( \bar{X}_1 - \bar{X}_2 \) = 3187 - 3214 = -27 miles. Then, calculate the standard error of the difference of means, which is \( SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \) = \sqrt{\frac{42.4^2}{45} + \frac{50.7^2}{40}} = 11.77 miles.
02

Construct a 95% confidence interval for the difference in means

For a 95% confidence interval, the critical z value is 1.96. The interval is given by \( (\bar{X}_1 - \bar{X}_2) \pm z(SE) \) = -27 \pm 1.96(11.77) = [-50.04, -3.96] miles.
03

Perform the hypothesis test

The null hypothesis is that the means are equal, or \( \mu_1 = \mu_2 \). The alternative hypothesis, however, is \( \mu_1 < \mu_2 \). Here we will use one-sided z-test. First, calculate the z-statistic: \( z = \frac{(\bar{X}_1 - \bar{X}_2) - 0}{SE} \) = -27 / 11.77 = -2.29. The p-value corresponding to this z value for a one-sided test is 0.011. Hence we reject the null hypothesis at the 1% level of significance because the p-value < 0.01. Thus, we can conclude that the mean distance between oil changes is less for all luxury cars than that for all compact lower-price cars.

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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 provides a range of values that is believed to contain a parameter of the population, such as the mean. It's an important aspect of inferential statistics that helps us understand data variability. In the context of the exercise, constructing a 95% confidence interval means we're 95% confident that the true difference in the mean distances between oil changes for luxury and compact cars lies within this range. We calculated it as
  • Lower bound: -50.04 miles
  • Upper bound: -3.96 miles
This interval tells us that, on average, the distance driven between oil changes is less for luxury cars than compact cars, as the interval is all negative. It indicates that following the oil change recommendations differs between luxury and compact car groups.
Z-Test
The z-test is a statistical method used to determine whether there is a significant difference between sample and population means, or between two sample means. This test is applicable when the population variance is known, or the sample size is large, typically over 30.
A one-sided z-test was performed in this exercise to see if luxury cars get oil changes at shorter intervals than compact cars. With the calculated z-statistic of -2.29 and corresponding p-value being less than 0.01, we find evidence to reject the null hypothesis. This suggests a statistically significant difference in oil change behavior, corroborating the hypothesis that luxury car drivers tend to adhere more closely to the oil change recommendations.
Significance Level
The significance level, often denoted as alpha (\(\alpha\), defines the cutoff for making decisions about the null hypothesis. A common choice in hypothesis testing is a 5% significance level, but in this exercise, a stricter 1% level was used.
This lower alpha means we require stronger evidence against the null hypothesis to conclude that luxury car drivers, on average, change oil more frequently. When the p-value of 0.011 was found, it was below the 1% significance threshold, leading to the conclusion that there’s a statistically significant difference in oil change patterns between the groups.
Sample Statistics
Sample statistics involve data from a sample of the population to estimate population parameters. It includes values like sample mean and standard deviation.
In our exercise, the samples came from 45 luxury car drivers and 40 compact car drivers, with means of 3187 and 3214 miles respectively. The goal was to estimate the population mean difference using these samples, leading us to confidently assert differences in behavior around oil changes. Understanding and calculating these statistics help make real-world decisions based on the data without needing to survey the entire population.

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Most popular questions from this chapter

A mail-order company has two warehouses, one on the West Coast and the second on the East Coast. The company's policy is to mail all orders placed with it within 72 hours. The company's quality control department checks quite often whether or not this policy is maintained at the two warehouses. A recently taken sample of 400 orders placed with the warehouse on the West Coast showed that 364 of them were mailed within 72 hours. Another sample of 300 orders placed with the warehouse on the East Coast showed that 279 of them were mailed within 72 hours.

We wish to estimate the difference between the mean scores on a standardized test of students taught by Instructors \(\mathrm{A}\) and \(\mathrm{B}\). The scores of all students taught by Instructor A have a normal distribution with a standard deviation of 15, and the scores of all students taught by Instructor B have a normal distribution with a standard deviation of 10 . To estimate the difference between the two means, you decide that the same number of students from each instructor's class should be observed. a. Assuming that the sample size is the same for each instructor's class, how large a sample should be taken from each class to estimate the difference between the mean scores of the two populations to within 5 points with \(90 \%\) confidence? b. Suppose that samples of the size computed in part a will be selected in order to test for the difference between the two population mean scores using a \(.05\) level of significance. How large does the difference between the two sample means have to be for you to conclude that the two population means are different? c. Explain why a paired-samples design would be inappropriate for comparing the scores of Instructor A versus Instructor \(\mathrm{B}\).

A company sent seven of its employees to attend a course in building self- confidence. These employees were evaluated for their self-confidence before and after attending this course. The following table gives the scores (on a scale of 1 to 15,1 being the lowest and 15 being the highest score) of these employees before and after they attended the course. $$ \begin{array}{l|rrrrrrr} \hline \text { Before } & 8 & 5 & 4 & 9 & 6 & 9 & 5 \\ \hline \text { After } & 10 & 8 & 5 & 11 & 6 & 7 & 9 \\ \hline \end{array} $$ a. Construct a \(95 \%\) confidence interval for the mean \(\mu_{d}\) of the population paired differences, where a paired difference is equal to the score of an employee before attending the course minus the score of the same employee after attending the course b. Test at the \(1 \%\) significance level whether attending this course increases the mean score of employees.

What is the shape of the sampling distribution of \(\hat{p}_{1}-\hat{p}_{2}\) for two large samples? What are the mean and standard deviation of this sampling distribution?

A company that has many department stores in the southern states wanted to find at two such stores the percentage of sales for which at least one of the items was returned. A sample of 800 sales randomly selected from Store A showed that for 280 of them at least one item was returned. Another sample of 900 sales randomly selected from Store B showed that for 279 of them at least one item was returned. a. Construct a \(98 \%\) confidence interval for the difference between the proportions of all sales at the two stores for which at least one item is returned. b. Using the \(1 \%\) significance level, can you conclude that the proportions of all sales for which at least one item is returned is higher for Store A than for Store \(B\) ?

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