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Chapter 9: Problem 70

Put on the Brakes The braking ability was compared for two 2012 automobile models. Random samples of 64 automobiles were tested for each type. The recorded measurement was the distance (in feet) required to stop when the brakes were applied at 50 miles per hour. These are the computed sample means and variances: Do the data provide sufficient evidence to indicate a difference between the mean stopping distances for the two models?

Short Answer

Expert verified
Answer: The null hypothesis (H0) states that there is no difference in the true mean stopping distances for the two models (渭1 - 渭2 = 0), while the alternative hypothesis (H1) states that there is a significant difference (渭1 - 渭2 鈮� 0).

Step by step solution

01

State the null and alternative hypotheses

The null hypothesis (H0) states that there is no difference in the true mean stopping distances for the two models, while the alternative hypothesis (H1) states that there is a significant difference. In mathematical terms: H0: 渭1 - 渭2 = 0, H1: 渭1 - 渭2 鈮� 0.
02

Calculate pooled variance and standard error

We want to calculate the pooled variance (sp^2) and standard error (SE) using the following formulas: sp^2 = ((n1-1)s1^2 + (n2-1)s2^2) / (n1 + n2 - 2) SE = sqrt((sp^2 / n1) + (sp^2 / n2)) Plug in the given values (n1=64, n2=64, s1^2, and s2^2) to obtain sp^2 and SE.
03

Calculate the test statistic

The test statistic (t) is calculated using the following formula: t = (x虅1 - x虅2) / SE Use the given sample means (x虅1 and x虅2) and the SE calculated in Step 2 to compute the test statistic.
04

Determine the degrees of freedom and find the critical value

The degrees of freedom for a two-sample t-test are: df = n1 + n2 - 2 For this exercise, df = 64 + 64 - 2 = 126. Now, let's choose a significance level (伪). If not specified, a common value to use is 伪 = 0.05 for a two-tailed test. Using a t-table or a calculator, find the critical t-value by consulting the row with df = 126 and 伪/2 = 0.025 (since we are doing a two-tailed test). This will give us the critical t-value.
05

Make a decision based on the test statistic and critical value

Compare the calculated test statistic (t) with the critical t-value. If the absolute value of the test statistic is greater than the critical t-value, we can reject the null hypothesis in favor of the alternative hypothesis, meaning that there is sufficient evidence to indicate a difference in the mean stopping distances between the two automobile models. If the absolute value of the test statistic is less than or equal to the critical t-value, we fail to reject the null hypothesis, meaning that there is not enough evidence to support a significant difference in the mean stopping distances.

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Suppose you wish to detect a difference between \(\mu_{1}\) and \(\mu_{2}\) (either \(\mu_{1}>\mu_{2}\) or \(\mu_{1}<\mu_{2}\) ) and, instead of running a two-tailed test using \(\alpha=.05,\) you use the following test procedure. You wait until you have collected the sample data and have calculated \(\bar{x}_{1}\) and \(\bar{x}_{2}\) If \(\bar{x}_{1}\) is larger than \(\bar{x}_{2},\) you choose the alternative hypothesis \(H_{\mathrm{a}}: \mu_{1}>\mu_{2}\) and run a one-tailed test placing \(\alpha_{1}=.05\) in the upper tail of the \(z\) distribution. If, on the other hand, \(\bar{x}_{2}\) is larger than \(\bar{x}_{1},\) you reverse the procedure and run a one-tailed test, placing \(\alpha_{2}=.05\) in the lower tail of the \(z\) distribution. If you use this procedure and if \(\mu_{1}\) actually equals \(\mu_{2},\) what is the probability \(\alpha\) that you will conclude that \(\mu_{1}\) is not equal to \(\mu_{2}\) (i.e., what is the probability \(\alpha\) that you will incorrectly reject \(H_{0}\) when \(H_{0}\) is true)? This exercise demonstrates why statistical tests should be formulated prior to observing the data.
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