91Ó°ÊÓ

In the early 1990's, researchers in the UK collected data on traffic flow, number of shoppers, and traffic accident related emergency room admissions on Friday the \(13^{\text {th }}\) and the previous Friday, Friday the \(6^{\text {th }}\). The histograms below show the distribution of number of cars passing by a specific intersection on Friday the \(6^{\text {th }}\) and Friday the \(13^{\text {th }}\) for many such date pairs. Also given are some sample statistics, where the difference is the number of cars on the 6 th minus the number of cars on the 13 th. \(^{20}\) $$\begin{array}{cccc} \hline & 6^{\text {th }} & 13^{\text {th }} & \text { Diff. } \\ \hline \bar{x} & 128,385 & 126,550 & 1,835 \\ s & 7,259 & 7,664 & 1,176 \\ n & 10 & 10 & 10 \\ \hline \end{array}$$ (a) Are there any underlying structures in these data that should be considered in an analysis? Explain. (b) What are the hypotheses for evaluating whether the number of people out on Friday the \(6^{\text {th }}\) is different than the number out on Friday the \(13^{\text {th }} ?\) (c) Check conditions to carry out the hypothesis test from part (b). (d) Calculate the test statistic and the p-value. (e) What is the conclusion of the hypothesis test? (f) Interpret the p-value in this context. (g) What type of error might have been made in the conclusion of your test? Explain.

Step by step solution

01

Identify the Structure in the Data

The data shows the distribution of car counts at an intersection on two Fridays. This is paired data across multiple Friday pairs. The difference in counts for each pair is calculated to analyze traffic patterns on Friday the 13th compared to the 6th.

02

Define the Hypotheses

For part (b), the null hypothesis (\(H_0\)) states that there is no difference in the mean number of cars, i.e., \(\mu_{\text{difference}} = 0\). The alternative hypothesis (\(H_a\)) is that there is a difference, i.e., \(\mu_{\text{difference}} eq 0\).

03

Check Conditions for Hypothesis Test

To perform the paired t-test, data must be approximately normally distributed or the sample size should be large. Here, since \(n = 10\) is small, we assume normality based on historical patterns or use Q-Q plots.

04

Calculate Test Statistic

Use the formula for the paired t-test statistic: \( t = \frac{\bar{x}_{\text{diff}}}{s_{\text{diff}}/\sqrt{n}} \). Plugging in the values, we get \( t = \frac{1835}{1176/\sqrt{10}} = 4.94 \).

05

Find the P-value

Using a t-distribution table, or software, find the p-value for \(t = 4.94\) with \(df = n-1 = 9\). This will be a small value, less than 0.05, indicating statistical significance.

06

State the Conclusion

Part (e) asks for the conclusion. Since the p-value is less than the significance level (typically 0.05), we reject the null hypothesis, indicating a significant difference in car counts between the two Fridays.

07

Interpret the P-value

The p-value indicates the probability of observing the test results, or something more extreme, assuming the null hypothesis is true. A small p-value suggests that the observed difference is unlikely under the null hypothesis.

08

Discuss Potential Errors

In hypothesis testing, a Type I error means rejecting a true null hypothesis. Since we rejected the null, there's a risk we made this error, i.e., concluding a difference when none exists.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Paired t-test

A paired t-test is a statistical procedure used to determine whether there is a significant difference between the means of two related groups. It's particularly useful when the data comes in pairs. In this exercise, the researchers are interested in examining if traffic flow differs significantly between two specific Fridays—the 6th and the 13th. When applying the paired t-test, the differences between each pair of observations are calculated.

• The calculation of differences helps to control for any variability or uniqueness of each pair, focusing solely on the change over time or condition.
• Hence, the variations between Fridays in the number of cars are highlighted through these differences.
• This approach helps in reducing the error variance that might occur if two unrelated samples were used instead.

It's essential that the differences follow a normal distribution, particularly if the sample size is small, like in the exercise with a sample size of 10. Ensuring this condition allows the paired t-test results to be valid and reliable.

Null Hypothesis

The null hypothesis (\( H_0 \)) is a foundational concept in hypothesis testing. It represents a statement of no effect or no difference. In this context, the null hypothesis posits that there is no difference in the average number of cars passing the intersection on the 6th compared to the 13th.Symbolically, this is expressed as:\(\mu_{\text{difference}} = 0\)The null hypothesis serves as a backdrop against which the alternative hypothesis is tested. It acts as a "starting assumption," assuming no difference or effect until evidence suggests otherwise. By concentrating on proving the null hypothesis wrong, researchers actually aim to find evidence that supports the alternative hypothesis.If the statistical analysis results in the null hypothesis being rejected, researchers conclude that there is a statistically significant effect or difference. If not, they fail to find evidence against it, leaving the null hypothesis as the best explanation given the data.

Alternative Hypothesis

The alternative hypothesis (\( H_a \)) is the counterpart to the null hypothesis. It suggests that there is an effect or a difference. In the case of this exercise, the alternative hypothesis would assert that the number of cars on the two Fridays, the 6th and the 13th, are different.Expressed in mathematical terms, this can be written as:\(\mu_{\text{difference}} eq 0\)Identifying the alternative hypothesis is crucial as it outlines the direction or nature of the expected effect. It indicates the specific interest of the hypothesis test—in this instance, determining if the unusual date of the 13th sees a reduction or increase in car numbers compared to the 6th.While testing hypotheses, the entire aim is to find significant evidence that suggests the alternative hypothesis should be adopted, refuting the null. Rejection of the null hypothesis in support of the alternative depends on the test results, especially focusing on statistics like the p-value, which indicate the likelihood of observing the data if the null were true.