It is advertised that the average braking distance for a small car traveling at 65 miles per hour equals 120 feet. A transportation researcher wants to determine if the statement made in the advertisement is false, meaning it is different than advertised. She randomly test drives 36 small cars at 65 miles per hour and records the braking distance. The sample average braking distance is computed as 126 feet. Assume that the population standard deviation is 22 feet.

State the null and alternative hypotheses. Using the critical value approach, test the hypotheses at the 5% level of significance.

Which of the following is the best conclusion to make on this problem?

Multiple Choice:

A. Since the calculated test statistic $ Z = 1.64 $ is equal to the critical value $-1.64$, at the $\alpha = 0.05$ level of significance we do not reject the null hypothesis.

B. Since the critical value $ +1.64 $ is greater than the calculated test statistic $ Z = -2.20 $, at the $\alpha = 0.05$ level of significance we reject the null hypothesis and conclude that the braking distance is less than 120.

C. Since the calculated test statistic $ Z = +1.64 $ is between the critical values $\pm 1.96$, at the $\alpha = 0.05$ level of significance we do not reject the null hypothesis.

D. Since the calculated test statistic $ Z = -1.64 $ is greater than the critical value $-1.96$, at the $\alpha = 0.05$ level of significance we accept the null hypothesis.

Question

01-08-2024
Mathematics

High School

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It is advertised that the average braking distance for a small car traveling at 65 miles per hour equals 120 feet. A transportation researcher wants to determine if the statement made in the advertisement is false, meaning it is different than advertised. She randomly test drives 36 small cars at 65 miles per hour and records the braking distance. The sample average braking distance is computed as 126 feet. Assume that the population standard deviation is 22 feet.

State the null and alternative hypotheses. Using the critical value approach, test the hypotheses at the 5% level of significance.

Which of the following is the best conclusion to make on this problem?

Multiple Choice:

A. Since the calculated test statistic $ Z = 1.64 $ is equal to the critical value $-1.64$, at the $\alpha = 0.05$ level of significance we do not reject the null hypothesis.

B. Since the critical value $ +1.64 $ is greater than the calculated test statistic $ Z = -2.20 $, at the $\alpha = 0.05$ level of significance we reject the null hypothesis and conclude that the braking distance is less than 120.

C. Since the calculated test statistic $ Z = +1.64 $ is between the critical values $\pm 1.96$, at the $\alpha = 0.05$ level of significance we do not reject the null hypothesis.

D. Since the calculated test statistic $ Z = -1.64 $ is greater than the critical value $-1.96$, at the $\alpha = 0.05$ level of significance we accept the null hypothesis.

Answer :

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LiamAlexanderSmith LiamAlexanderSmith · Answer 1 · 2024-12-11T08:49:47-05:00

In this problem, we need to determine if the transportation researcher's claim about the braking distance differs from what is advertised — 120 feet for a small car traveling at 65 miles per hour. This is done using hypothesis testing. Let's break this down step-by-step:

1. Stating the Hypotheses:

The hypotheses for this test can be formulated as:

Null Hypothesis ([tex]H_0[/tex]): The average braking distance is equal to 120 feet. Mathematically, this is [tex]H_0: \mu = 120[/tex].
Alternative Hypothesis ([tex]H_a[/tex]): The average braking distance is not equal to 120 feet. Mathematically, this is [tex]H_a: \mu \neq 120[/tex].

This is a two-tailed test because we are testing if the braking distance is different (either greater or less) than 120 feet.

2. Critical Value Approach:

The critical values for a two-tailed test at a 5% significance level ([tex]\alpha = 0.05[/tex]) can be found using a standard normal distribution (Z-distribution). Based on Z-tables, the critical Z-values are approximately [tex]\pm 1.96[/tex]. This means the rejection region for the null hypothesis falls outside of this range.

3. Calculating the Test Statistic:

We use the formula for the Z-test statistic for the sample mean:

[tex]Z = \frac{\bar{x} - \mu_0}{\sigma/\sqrt{n}}[/tex]

Where:

[tex]\bar{x} = 126[/tex] (sample mean)
[tex]\mu_0 = 120[/tex] (mean advertised)
[tex]\sigma = 22[/tex] (population standard deviation)
[tex]n = 36[/tex] (sample size)

Plugging in the values, we get:

[tex]Z = \frac{126 - 120}{22/\sqrt{36}} = \frac{6}{3.667} \approx 1.64[/tex]

4. Conclusion:

The calculated Z-statistic is approximately 1.64.

Comparing this to the critical Z-values:

Since $1.64[tex]is between[/tex]-1.96[tex]and $1.96[/tex], we do not reject the null hypothesis.

Thus, the conclusion to make in this scenario, from the given multiple choice options, is:

"Since the calculated test statistic Z=+1.64 is between the critical values +/- 1.96, at the [tex]\alpha=0.05[/tex] level of significance we do not reject the null hypothesis."