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Chapter 9: Problem 19

Suppose \(\mu_{\mathrm{t}}\) and \(\mu_{2}\) are true mean stopping distances at 50 mph for cars of a certain type equipped with two different types of braking systems. Use the two-sample \(t\) test at sig" nificance level .01 to test \(H_{0}=\mu_{1}-\mu_{2}=-10\) versus \(H_{2}\) : \(\mu_{4}-\mu_{2}<-10\) for the following dataz \(m=6, \bar{x}=115.7\), \(s_{1}=5.03, n-6, \bar{y}=129.3\), and \(s_{2}=5.38 .\)

Short Answer

Expert verified
Do not reject \( H_0 \); insufficient evidence for \( \mu_1 - \mu_2 < -10 \).

Step by step solution

01

Clarify the Hypotheses

Given the problem, we need to clarify that the null hypothesis, \( H_0 \), is \( \mu_1 - \mu_2 = -10 \). The alternative hypothesis, \( H_a \), is \( \mu_1 - \mu_2 < -10 \). We are testing if the difference in means is less than -10.
02

Gather the Sample Information

We have the sample size, mean, and standard deviation for both groups: \( m = 6, \bar{x} = 115.7, s_1 = 5.03 \) for the first group, and \( n = 6, \bar{y} = 129.3, s_2 = 5.38 \) for the second group.
03

Calculate the Test Statistic

Use the formula for the test statistic for a difference in means:\[t = \frac{(\bar{x} - \bar{y}) - (-10)}{\sqrt{\frac{s_1^2}{m} + \frac{s_2^2}{n}}} = \frac{(115.7 - 129.3) + 10}{\sqrt{\frac{5.03^2}{6} + \frac{5.38^2}{6}}}.\]First calculate the pooled standard deviation and then substitute to find \( t \).
04

Calculate the Pooled Standard Deviation

First, compute the variances: \[\frac{s_1^2}{m} = \frac{5.03^2}{6}, \quad \frac{s_2^2}{n} = \frac{5.38^2}{6}.\]Thus,\[\sqrt{\frac{5.03^2}{6} + \frac{5.38^2}{6}} \approx \sqrt{4.216472 + 4.828173}" \approx \sqrt{9.044645} \approx 3.006.\]
05

Compute the Test Statistic

Using the pooled standard deviation from Step 4, calculate \( t \):\[t = \frac{(115.7 - 129.3) + 10}{3.006} = \frac{-13.6 + 10}{3.006} = \frac{-3.6}{3.006} \approx -1.198.\]
06

Determine the Critical Value

For a two-sample t-test with \( m + n - 2 = 10 \) degrees of freedom at a significance level of \( \alpha = 0.01 \), a one-sided critical value can be found using t-distribution tables or technology, which is approximately \(-2.764\).
07

Compare Statistic to Critical Value

Our calculated test statistic \( t \approx -1.198 \) is greater than our critical value of \(-2.764\). Thus, we do not reject \( H_0 \).
08

Conclusion

Since \( t \approx -1.198 \) is greater than the critical value, there is insufficient evidence to support the claim that the mean difference in stopping distance is less than -10.

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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 foundational concept in statistics that helps us determine if there is enough evidence in a sample of data to infer that a certain condition holds true in the entire population. In this exercise, we have two hypotheses:

  • The null hypothesis (\( H_0 \)) suggests no significant change, stated as \( \mu_1 - \mu_2 = -10 \).
  • The alternative hypothesis (\( H_a \)) challenges the null, indicating that the difference in means is less than -10, or \( \mu_1 - \mu_2 < -10 \)

In hypothesis testing, our goal is to weigh the evidence provided by the sample data against the initial assumption of the null hypothesis. If the evidence is strong enough, we reject the null hypothesis in favor of the alternative one.
T-Distribution
The t-distribution is an essential tool used in statistics for hypothesis testing, especially when dealing with small sample sizes, typically fewer than 30 observations. It looks similar to the normal distribution but has heavier tails, which means it reflects more variability and accounts for uncertainties better with smaller samples.

This exercise involves a two-sample t-test, where we compare the means of two groups. Because the sample sizes are small (\( m = 6 \) and \( n = 6 \)), we use the t-distribution rather than the normal distribution to determine the test statistic and critical values.
Test Statistic
The test statistic is a key figure calculated from the sample data and used in hypothesis testing to help decide whether to reject the null hypothesis. We compute it by using the given formula:

\[ t = \frac{(\bar{x} - \bar{y}) - (-10)}{\sqrt{\frac{s_1^2}{m} + \frac{s_2^2}{n}}} \]

This formula helps us understand how many standard deviations our sample difference is from the null hypothesis difference (-10 in this case). In the given problem, the computed test statistic is approximately -1.198. The result helps us compare this with the critical value to draw conclusions about the hypotheses.
Critical Value
The critical value is a threshold that our test statistic is compared against to decide whether to reject or accept the null hypothesis. It is determined based on the significance level and the degrees of freedom in our data.

In this exercise, the significance level \( \alpha \) is set at 0.01, and the degrees of freedom calculated as \( m + n - 2 = 10 \). Looking up a t-distribution table or using statistical software, a one-tailed critical value is approximately -2.764.

The decision rule is straightforward: if the test statistic is less than or equal to the critical value, we reject the null hypothesis. However, in this case, since our calculated t-value (-1.198) is greater than the critical value (-2.764), we do not have sufficient evidence to reject the null hypothesis, so it remains as is.

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Most popular questions from this chapter

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