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Chapter 10: Problem 9

A car magazine is comparing the total repair costs incurred during the first three years on two sports cars, the T-999 and the XPY. Random samples of 45 T-999s and 51 XPYs are taken. All 96 cars are 3 years old and have similar mileages. The mean of repair costs for the 45 T-999 cars is \(\$ 3300\) for the first 3 years. For the 51 XPY cars, this mean is \(\$ 3850\). Assume that the standard deviations for the two populations are \(\$ 800\) and \(\$ 1000\), respectively. a. Construct a 99\% confidence interval for the difference between the two population means. b. Using a \(1 \%\) significance level, can you conclude that such mean repair costs are different for these two types of cars? c. What would your decision be in part b if the probability of making a Type I error were zero? Explain.

Short Answer

Expert verified
a. The 99% confidence interval for the difference between the two population means is -$550 卤 $888. b. At a 1% significance level, there is no evidence to conclude that the mean repair costs are different for the two types of cars. c. If the probability of making a Type I error were zero, the decision would remain the same, as there's no evidence that the mean repair cost of the two cars are different.

Step by step solution

01

Calculate The Confidence Interval

Here, the goal is to find a 99% confidence interval for the difference between the population means. The formula to calculate a confidence interval for the difference between two means (assuming the standard deviations are known) is: \[\bar{X1} - \bar{X2} \pm Z \cdot \sqrt{\frac{{S1}^2}{n1} + \frac{{S2}^2}{n2}}\]Where, \(\bar{X1}\) = mean of population 1 = $3300\(\bar{X2}\) = mean of population 2 = $3850\(S1\) = standard deviation of population 1 = $800\(n1\) = size of the sample 1 = 45\(S2\) = standard deviation of population 2 = $1000\(n2\) = size of the sample 2 = 51Z = Z-score for desired confidence level 99% = 2.33 (from Z-table)Substituting all values, the 99% confidence interval will be ($3300-$3850) 卤 (2.33* \(\sqrt{(\frac{800^2}{45} + \frac{1000^2}{51})}\)) = -$550 卤 (2.33* \(\sqrt{(\frac{640000}{45} + \frac{1000000}{51})}\)) = -$550 卤 $888.
02

Make the Hypothesis Test

We perform a hypothesis test to determine if the mean repair costs for the two types of cars are statistically different at a 1% significance level. For this, we use the following hypotheses:\(H0\): The mean repair costs are the same for the two types of cars (\(\mu1 - \mu2\) = 0)\(HA\): The mean repair costs are different for the two types of cars (\(\mu1 - \mu2\) 鈮� 0)The rejection region for the 1% level test is z < -2.33 or z > 2.33. The test statistic is calculated as:Z = (\(\bar{X1} - \bar{X2}\) - (\(\mu1 - \mu2\)) / \(\sqrt{\frac{{S1}^2}{n1} + \frac{{S2}^2}{n2}}\)Substituting the values, we have:Z = ($3300-$3850) / \(\sqrt{(\frac{800^2}{45} + \frac{1000^2}{51})}\) = -1.91Since -1.91 does not fall into the rejection region (-2.33, 2.33), we cannot reject the null hypothesis. Therefore, there's no evidence that the mean repair cost of the two cars are different.
03

Make Decision If Type I Error Were Zero

A Type I error occurs when the null hypothesis is rejected when it is actually true. If the probability of making a Type I error were zero, it would mean that there would be no chances of incorrectly rejecting the null hypothesis. Therefore, there would be no chances of incorrectly concluding that the mean repair costs are different for these two types of cars. Thus, the decision in part b would remain the same: there's no evidence that the mean repair cost of the two cars are different.

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Key Concepts

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

Confidence Interval
When we talk about confidence intervals, we're looking at a range that estimates where something like a population mean might fall. Imagine you have measured something, like repair costs for cars. Instead of saying "this is exactly the cost," we say, "we're 99% confident the true mean cost difference is in this range." This helps account for random variation in data.

A 99% confidence interval is particularly wide to ensure a high degree of certainty, but it still remains a guess based on sample data and the assumption of normality. The formula used helps us estimate how far our sample mean might be from the actual population mean, and factors in variations from the respective populations.
  • The bigger your sample size, the more precise your interval will be.
  • Need to know what's happening across the whole population? That鈥檚 where the confidence interval can step in!
Population Means
The concept of population means is essential in understanding the central tendency of a population鈥檚 data. A population mean is the average of a set of data that includes every member of that population. For example, when comparing two different sports cars, determining the mean repair cost provides a clear indication of what owners can generally expect.

We calculate this mean by summing up all observed values and dividing by the total number of observations. In the exercise, we look at means for both the T-999 and XPY cars. These means tell us, on average, how much repair costs are over a certain period.
  • Population means provide a snapshot of average costs across all individuals in the population.
  • Sampling can offer a practical method to estimate these means without surveying all individuals.
Standard Deviation
Standard deviation is a measure of how spread out the numbers in your data set are. A lower standard deviation means that values tend to be close to the mean, while a higher standard deviation indicates that the values are more spread out.

Think of it as a measure of dependability in your estimates. If we're considering repair costs, a low standard deviation suggests that most cars are likely having similar repair costs. The formula used involves subtracting each value from the mean, squaring that result, finding the average of those squared differences, and then taking the square root.
  • It helps determine if the data is closely packed or widely dispersed.
  • In our exercise, each car brand has a specific standard deviation that shows cost variability within samples.
Type I Error
A Type I error occurs in hypothesis testing when a true null hypothesis is rejected. Essentially, it's like saying there's an effect when there isn't. Imagine you wrongly tell everyone that a new flavor of cake is yucky when it's actually delicious.

In statistical terms, this is akin to crying wolf when there isn't one, based on our sample data. Being mindful of Type I errors is crucial, especially when the stakes are high, as we falsely conclude that a difference or effect exists.
  • Minimizing Type I errors is key to making reliable conclusions.
  • In our car repair cost example, avoiding this error means not hastily claiming costs differences when none statistically exists.

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Most popular questions from this chapter

When are the samples considered large enough for the sampling distribution of the difference between two sample proportions to be (approximately) normal?

A high school counselor wanted to know if tenth-graders at her high school tend to have the same free time as the twelfth-graders. She took random samples of 25 tenth-graders and 23 twelfth-graders. Each student was asked to record the amount of free time he or she had in a typical week. The mean for the tenth-graders was found to be 29 hours of free time per week with a standard deviation of \(7.0\) hours. For the twelfth-graders, the mean was 22 hours of free time per week with a standard deviation of \(6.2\) hours. Assume that the two populations are approximately normally distributed with unknown but equal standard deviations. a. Make a \(90 \%\) confidence interval for the difference between the corresponding population means. b. Test at a \(5 \%\) significance level whether the two population means are different.

The Pew Research Center conducted a poll in January 2014 of online adults who use social networking sites. According to this poll, \(89 \%\) of the \(18-29\) year olds and \(82 \%\) of the \(30-49\) year olds who are online use social networking sites (www.pewinternet.org). Suppose that this survey included 562 online adults in the \(18-29\) age group and 624 in the \(30-49\) age group. a. Let \(p_{1}\) and \(p_{2}\) be the proportion of all online adults in the age groups \(18-29\) and \(30-49\), respectively, who use social networking sites. Construct a \(95 \%\) confidence interval for \(p_{1}-p_{2}\) b. Using a \(1 \%\) significance level, can you conclude that \(p_{1}\) is different from \(p_{2} ?\) Use both the critical-value and the \(p\) -value approaches.

Maria and Ellen both specialize in throwing the javelin. Maria throws the javelin a mean distance of 200 feet with a standard deviation of 10 feet, whereas Ellen throws the javelin a mean distance of 210 feet with a standard deviation of 12 feet. Assume that the distances each of these athletes throws the javelin are normally distributed with these population means and standard deviations. If Maria and Ellen each throw the javelin once, what is the probability that Maria's throw is longer than Ellen's?

According to a Bureau of Labor Statistics report released on March 25,2015 , statisticians earn an average of \(\$ 84,010\) a year and accountants and auditors earn an average of \(\$ 73,670\) a year (www.bls. gov). Suppose that these estimates are based on random samples of 2000 statisticians and 1800 accountants and auditors. Further, assume that the sample standard deviations of the annual earnings of these two groups are \(\$ 15,200\) and \(\$ 14,500\), respectively, and the population standard deviations are unknown but equal for the two groups. a. Construct a \(98 \%\) confidence interval for the difference in the mean annual earnings of the two groups, statisticians and accountants and auditors. b. Using a \(1 \%\) significance level, can you conclude that the average annual earnings of statisticians is higher than that of accountants and auditors?
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