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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 XPY's 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 3years. 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
To definitively answer part a, b, and c, we need to perform the calculations above. However, generally speaking, the existence of a Type I error changes the conclusion of our hypothesis test, because it alters the threshold for rejecting the null hypothesis.

Step by step solution

01

Define the Variables and Hypotheses

Firstly, let's define the data given and the hypotheses for the test:\n\n\(X\): Mean repair cost of T-999, \(\mu1 = 3300\), \(s1 = 800\), \(n1 = 45\)\n\(Y\): Mean repair cost of XPY, \(\mu2 = 3850\), \(s2 = 1000\), \(n2 = 51\)\n\nThe hypotheses for the test are:\n\(H0\): \(\mu1 - \mu2 = 0\) (Mean repair costs are equal)\n\(H1\): \(\mu1 - \mu2 \neq 0\) (Mean repair costs are different)
02

Construct the Confidence Interval

A 99% confidence interval for the difference in population means can be calculated using the following formula:\n\n\(\[ \mu1 - \mu2 \pm z \left( \sqrt{ \frac{s1^2}{n1} + \frac{s2^2}{n2} } \right) \]\).\n\nFor a 99% confidence interval, z-value is 2.58. Substituting the given values into the formula, the interval can be calculated.
03

Interpret the Confidence Interval

The resulting interval gives an estimate of the difference in population means. If this interval contains 0, it means that the difference in repair costs is not statistically significant at the 99% confidence level.
04

Perform the Hypothesis Test

We can perform a two-sample z-test to test the hypotheses at a 1% significance level. The z-score for the sample data difference is calculated as follows:\n\n\[ Z = \frac{(Xbar - Ybar) - D0}{\sqrt{ \frac{s1^2}{n1} + \frac{s2^2}{n2} } }\]\n\nSubstituting the data, we calculate the z-score and we compare it with critical z-value for the 1% level of significance (z = 2.58).
05

Make Decision

If the calculated z-value is greater than 2.58 (in absolute value), we reject the null hypothesis, else, we do not reject. It means that if we reject the null hypothesis, we can state that there is evidence that the mean repair costs are different for the two types of cars at a 1% significance level.
06

Adjusting for Type I Error

If the probability of making Type I error were zero, it would imply that we are 100% sure rejecting a true null hypothesis, i.e., incorrectly stating that means are different when actually they're equal.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a statistical method used to make decisions about the properties of populations based on sample data. It's like setting up a trial where you start with a claim (null hypothesis) and test it using evidence (data). In this scenario, we want to determine if there's a significant difference in repair costs between two car models, the T-999 and the XPY.

To do this, we set up two hypotheses:
  • **Null Hypothesis (\(H_0\))**: This assumes that the average repair costs for the T-999 and XPY are the same, meaning any observed difference is due to random chance. Mathematically, it’s \(\mu_1 - \mu_2 = 0\).
  • **Alternative Hypothesis (\(H_1\))**: This suggests that there is a real difference in repair costs between the two car models, expressed as \(\mu_1 - \mu_2 eq 0\).
By conducting a hypothesis test, we evaluate these hypotheses using our sample data. If the test provides sufficient evidence, we might reject the null hypothesis in favor of the alternative.
Two-Sample Z-Test
The two-sample z-test is perfect for comparing the means of two different populations, especially when dealing with larger sample sizes and known population standard deviations. Here's how it works in the context of our car repair cost comparison:

  • Calculate the difference between the sample means: \(X_{bar} - Y_{bar} = 3300 - 3850 = -550\).
  • Determine the standard error using the formula \(\sqrt{ \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }\) where \(s_1 = 800\) and \(s_2 = 1000\), with \(n_1 = 45\) and \(n_2 = 51\).
  • Calculate the z-score: \[ Z = \frac{(X_{bar} - Y_{bar}) - 0}{\text{Standard Error}} \]. This z-score tells us how many standard deviations the observed difference is from the expected zero difference under the null hypothesis.
We then compare this z-score to the critical value at the desired significance level (1% here, \(z = 2.58\) for a two-tailed test). If the absolute value of the z-score is larger than the critical value, we reject the null hypothesis.
Type I Error
In the world of hypothesis testing, a Type I error is a false positive. It occurs when we wrongly reject a true null hypothesis. Think of it as seeing a difference when none actually exists. The significance level (\(\alpha\)) helps control this error. For example, a 1% significance level means we're willing to accept a 1% chance of making this error.

In the context of comparing car repair costs, if we set our significance level at 1%, we are intentionally limiting our chance of making a Type I error to 1%. This means there's only a 1% chance we’ll mistakenly conclude that repair costs differ between the two car models when they don’t.

If the chance of a Type I error were zero, that would mean we'd be making absolutely no mistakes in detecting differences. However, in practical scenarios, we accept a small chance to ensure our conclusions are robust, allowing us to be reasonably confident in our decision to support the alternative hypothesis.

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Most popular questions from this chapter

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