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The standard recommendation for automobile oil changes is once every 5000 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 lowerprice cars were selected. The average distance driven between oil changes was 5187 miles for the luxury car owners and 5214 miles for the compact lower-price cars. The sample standard deviations were 424 and 507 miles for the luxury and compact groups, respectively. Assume that the two population distributions of the distances between oil changes have the same standard deviation. 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 a \(1 \%\) 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?

Step by step solution

01

State the hypotheses

The null hypothesis \(H_0\) is that the means are equal: \(\mu_{luxury} = \mu_{compact}\). The alternate hypothesis \(H_a\) is that mean of luxury cars is less than compact cars: \(\mu_{luxury} < \mu_{compact}\).

02

Calculate the combined sample standard deviation

We need to calculate the pooled standard deviation given by \(s_p = \sqrt{( (n1-1)*s1^2 + (n2-1)*s2^2 ) / (n1+n2-2)}\), where \(n1\), \(n2\) are the sample sizes and \(s1\), \(s2\) are the standard deviations of the two groups. In this case, \(s_p = \sqrt{((45-1)*424^2 + (40-1)*507^2) / (45+40-2)} = 465.71 miles.\)

03

Calculate the confidence interval

The confidence interval is calculated using the formula \((\bar{x}_1 - \bar{x}_2) \pm (t_{\alpha/2}*s_p\sqrt{1/n_1 + 1/n_2})\) where \(\(\bar{x}_1 - \bar{x}_2\) is the difference between the two sample means, \(t_{\alpha/2}\) is the t-value corresponding to the confidence level \(95\%\) with degree of freedom equals \(n_1 + n_2 - 2\). In this case, our confidence interval would be \((5187 - 5214) \pm (2.024 * 465.71 * \sqrt{1/45 + 1/40}) = -27 \pm 138.71\) miles. Therefore, the 95% confidence interval is between -165.71 and 111.71.

04

Perform hypothesis testing

To perform hypothesis testing, we need to calculate the t score using the formula \(t = (\bar{x}_1 - \bar{x}_2) / (s_p\sqrt{1/n_1 + 1/n_2})\). In this case, \(t = (5187 - 5214) / (465.71 *\sqrt{1/45 + 1/40}) = -0.1789\). Compare this t-value with the critical t-value corresponding to the 1% significance level (which is -2.626 given \(df = n1+n2-2\)). Since -0.1789 > -2.626, we cannot reject the null hypothesis at 1% significance level, meaning there's not enough statistical evidence to show luxury car owners change their oil less frequently than compact car owners.

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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 is a range of values, derived from the sample data, that is likely to contain the true population parameter. In hypothesis testing, it allows us to make predictions about population parameters with a certain degree of assurance. Calculating a 95% confidence interval tells us that we can be 95% confident that the interval contains the true difference in means between the two groups being compared.
In the context of our exercise, this involves finding the mean difference in miles between oil changes for luxury and compact cars and incorporating the variability of both samples. The equation used for calculating the confidence interval is:

• \[ (ar{x}_1 - ar{x}_2) \pm (t_{\alpha/2} \cdot s_p \sqrt{1/n_1 + 1/n_2}) \]

This equation includes the sample means, the pooled standard deviation \((s_p)\), and the critical t-value.
For our data, the calculated confidence interval was between -165.71 and 111.71 miles, indicating that the true mean difference in miles between oil changes could fall within this range.

T-Distribution

The t-distribution is a probability distribution that is used in statistical analysis to estimate population parameters when the sample size is small and/or the population variance is unknown. It is similar to the normal distribution but with thicker tails, allowing for the increased uncertainty in estimates from smaller samples.
In this problem, the t-distribution is used to find the critical t-value, which helps in constructing the confidence interval and performing hypothesis testing. The t-value of \(t_{\alpha/2}\) used in our confidence interval calculation depends on the desired confidence level and the degrees of freedom (which, in this case, is the total number of samples minus two, \( df = n_1 + n_2 - 2 \)).
For the 95% confidence level with 83 degrees of freedom, the critical t-value was approximately 2.024. This value is utilized to determine the margin of error for our interval.

Pooled Standard Deviation

The pooled standard deviation is a method of estimating the standard deviation of two independent samples assuming they have equal population variances. It represents a way to combine the variances of the two groups into a single measure.
To calculate the pooled standard deviation, use the formula:

• \[ s_p = \sqrt{\frac{ (n_1-1)s_1^2 + (n_2-1)s_2^2 }{n_1 + n_2 - 2}} \]

Here, \( s_p \) is the pooled standard deviation, \( s_1 \) and \( s_2 \) are the sample standard deviations, and \( n_1 \) and \( n_2 \) are the sample sizes for each group.
In our exercise, the pooled standard deviation calculated was 465.71 miles. This common standard deviation becomes crucial when comparing the means of the two groups, as it allows for a more accurate assessment of the variability across groups.

Significance Level

The significance level, often denoted as \( \alpha \), is the probability of rejecting the null hypothesis when it is actually true. It represents the risk we take of making a Type I error in statistical hypothesis testing.
In practice, a common significance level used is \( \alpha = 0.05 \), but in this problem, a stricter \( \alpha = 0.01 \) is used. This means that there is only a 1% risk of wrongly rejecting the null hypothesis.
The lower the significance level, the stronger the evidence must be to reject the null hypothesis. In our case, the calculated t-value from the test was compared against the critical t-value from the t-distribution table at the 1% level, which helps to decide whether to support or reject the null hypothesis.