/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 13 A car magazine is comparing the ... [FREE SOLUTION] | 91Ó°ÊÓ

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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 the \(1 \%\) 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 \(\mathrm{b}\) if the probability of making a Type I error were zero? Explain.

Short Answer

Expert verified
Step 1 will yield a specific 99% confidence interval for the difference between the population means. The action in Step 2 fully depends on the calculated P-value and the given significance level (\(alpha\)). Step 3 describes that with zero probability of a Type I error, maximum confidence in the rejection of the null hypothesis would be required, which is practically unachievable in real-life scenarios.

Step by step solution

01

Construct a 99% confidence interval

First, calculate the standard error of the difference between the two sample means using the formula \(\sqrt{(\sigma_1^2/n_1) + (\sigma_2^2/n_2)}\), where \(\sigma_1\) and \(\sigma_2\) are the standard deviations and \(n_1\) and \(n_2\) are the respective sample sizes. The sample means are denoted as \(\overline{x_1}\) and \(\overline{x_2}\). Then, we find the critical value \[Z_{(1 + (1 - 0.99)/2)}\] from the Z-table. The confidence interval can then be created as \[(\overline{x_1} - \overline{x_2}) \pm Z_{(1 + (1 - 0.99)/2)} \cdot SE\]
02

Hypothesis testing

Set up the null hypothesis (\(H_0\)) and the alternative hypothesis (\(H_1\)) as follows: \(H_0: \mu_1 - \mu_2 = 0\) (\(i.e.,\ there is no difference in the mean repair costs between T-999 and XPY\)) and \(H_1: \mu_1 - \mu_2 \neq 0\) (\(i.e.,\ there is a difference in mean repair costs between T-999 and XPY\)). Now, calculate the Z statistic using the formula \[Z = (\overline{x_1} - \overline{x_2}) - 0 / SE\] Then, find the P-value associated with the calculated Z score using the Z-table, and compare it with the significance level (\(\alpha= 0.01\)). If the P-value is less than 0.01, we reject the null hypothesis, otherwise we fail to reject the null hypothesis.
03

Discussion of a Type I error

Recall that a Type I error occurs when we reject the null hypothesis when it is true. If the probability of making a Type I error was zero, this would mean that we are perfectly certain that we would not wrongly reject the null hypothesis. In other words, if we rejected the null hypothesis, we would have 100% certainty that the average repair costs for T-999 and XPY are indeed different. In reality, achieving zero probability of a Type I error is virtually impossible without exhaustive information about the population, and tests typically involve balancing Type I and Type II errors.

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Key Concepts

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

Understanding Confidence Intervals
When we talk about a confidence interval, we refer to a range that estimates the difference between two population means. This range tells us how confident we are that the true difference falls within it. In our exercise, we're looking at the repair costs for two car models. To construct the confidence interval, we calculate the standard error, a simple measure of how accurately our sample represents the population. The formula for the standard error involves the standard deviations and sample sizes of the two groups. After finding this, we look up the critical value from the Z-table, based on our desired confidence level—99% in this case. The confidence interval is then built around the difference in sample means, adding and subtracting the product of the critical value and standard error. This interval gives us a range where the true difference in repair costs is likely to fall 99% of the time if we repeated this study with new samples.
Principles of Hypothesis Testing
Hypothesis testing is a method used to make decisions about population parameters. We start with a null hypothesis (\(H_0\)), which suggests no effect or no difference. For our exercise, \(H_0: \mu_1 - \mu_2 = 0\), means there's no difference in the average repair costs of the two cars. The alternative hypothesis (\(H_1\)) suggests that there is a difference. We use the Z-test to determine which hypothesis the data supports. This involves calculating a Z statistic using the difference in sample means and dividing it by the standard error. Once we have the Z statistic, we check its corresponding probability, known as the p-value. If the p-value is less than the chosen significance level (\(\alpha = 0.01\)), it suggests that the observed data would be very unlikely under the null hypothesis, leading us to reject \(H_0\).
Understanding Type I Error
Type I error is a crucial idea in statistics. It's when we falsely reject a true null hypothesis. Imagine if we conclude that the two cars have different repair costs when, in fact, they don't—that's a Type I error. In testing, this error is controlled by setting a significance level (\(\alpha\)), typically at 0.05 or 0.01, like in our exercise. This significance level is the probability of making a Type I error, meaning that 1% of the time, we might incorrectly reject the null hypothesis if our \(\alpha\) is 0.01. If the probability of making a Type I error were zero, it would mean we never make a false claim of difference. However, this is practically impossible as it would require perfect knowledge of the populations.
Conducting a Z-Test
A Z-test is a statistical test used to determine differences between sample means. It's applicable when we know the population standard deviations, as in our exercise. To perform a Z-test, we first state our null and alternative hypotheses. Then, we calculate the Z statistic, which measures how far away our sample statistic is from the null hypothesis value—in standard errors. The formula for the Z statistic is \(Z = \frac{(\overline{x_1} - \overline{x_2})}{SE}\), where \(SE\) is the standard error. Next, we compare the calculated Z statistic against the critical Z value from a Z-table. If the calculated Z statistic is greater than the critical value or the p-value is less than our significance level, we reject the null hypothesis. This tells us that there is significant evidence to suggest a difference in mean repair costs between the two car models.

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Most popular questions from this chapter

Manufacturers of two competing automobile models, Gofer and Diplomat, each claim to have the lowest mean fuel consumption. Let \(\mu_{1}\) be the mean fuel consumption in miles per gallon (mpg) for the Gofer and \(\mu_{2}\) the mean fuel consumption in mpg for the Diplomat. The two manufacturers have agreed to a test in which several cars of each model will be driven on a 100 -mile test run. Then the fuel consumption, in mpg, will be calculated for each test run. The average of the mpg for all 100 -mile test runs for each model gives the corresponding mean. Assume that for each model the gas mileages for the test runs are normally distributed with \(\sigma=2 \mathrm{mpg}\). Note that each car is driven for one and only one 100 -mile test run. a. How many cars (i.e., sample size) for each model are required to estimate \(\mu_{1}-\mu_{2}\) with a \(90 \%\) confidence level and with a margin of error of estimate of \(1.5 \mathrm{mpg}\) ? Use the same number of cars (i.e., sample size) for each model. b. If \(\mu_{1}\) is actually \(33 \mathrm{mpg}\) and \(\mu_{2}\) is actually \(30 \mathrm{mpg}\), what is the probability that five cars for each model would yield \(\bar{x}_{1} \geq \bar{x}_{2}\) ?

Explain when you would use the paired-samples procedure to make confidence intervals and test hypotheses.

A sample of 500 observations taken from the first population gave \(x_{1}=305\). Another sample of 600 observations taken from the second population gave \(x_{2}=348\). a. Find the point estimate of \(p_{1}-p_{2}\). b. Make a \(97 \%\) confidence interval for \(p_{1}-p_{2}\). c. Show the rejection and nonrejection regions on the sampling distribution of \(\hat{p}_{1}-\hat{p}_{2}\) for \(H_{0}: p_{1}=p_{2}\) versus \(H_{1}: p_{1}>p_{2} .\) Use a significance level of \(2.5 \% .\) d. Find the value of the test statistic \(z\) for the test of part \(\mathrm{c}\). e. Will you reject the null hypothesis mentioned in part \(\mathrm{c}\) at a significance level of \(2.5 \%\) ?

The following information was obtained from two independent samples selected from two normally distributed populations with unknown but equal standard deviations. $$ \begin{array}{llllllllllllll} \text { Sample 1: } & 47.7 & 46.9 & 51.9 & 34.1 & 65.8 & 61.5 & 50.2 & 40.8 & 53.1 & 46.1 & 47.9 & 45.7 & 49.0 \\ \text { Sample 2: } & 50.0 & 47.4 & 32.7 & 48.8 & 54.0 & 46.3 & 42.5 & 40.8 & 39.0 & 68.2 & 48.5 & 41.8 & \end{array} $$ a. Let \(\mu_{1}\) be the mean of population 1 and \(\mu_{2}\) be the mean of population \(2 .\) What is the point estimate of \(\mu_{1}-\mu_{2} ?\) b. Construct a \(98 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). c. Test at the \(1 \%\) significance level if \(\mu_{1}\) is greater than \(\mu_{2}\).

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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