/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 73 Every road has one at some point... [FREE SOLUTION] | 91Ó°ÊÓ

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Every road has one at some point - construction zones that have much lower speed limits. To see if drivers obey these lower speed limits, a police officer uses a radar gun to measure the speed (in miles per hours, or mph) of a random sample of 10 drivers in a 25 mph construction zone. Here are the data: \(\begin{array}{llllllllll}27 & 33 & 32 & 21 & 30 & 30 & 29 & 25 & 27 & 34\end{array}\) (a) Is there convincing evidence that the average speed of drivers in this construction zone is greater than the posted speed limit? (b) Given your conclusion in part (a), which kind of mistake - a Type I error or a Type II error - could you have made? Explain what this mistake would mean in context.

Short Answer

Expert verified
(a) There is evidence that the average speed is greater than 25 mph. (b) A Type I error could occur, which means concluding speeds exceed 25 mph when they do not.

Step by step solution

01

Define the Hypotheses

The first step is to define the null and the alternative hypothesis. Here, the null hypothesis \( H_0 \) is that the average speed of drivers in the construction zone is equal to 25 mph. So, \( H_0: \mu = 25 \). The alternative hypothesis \( H_a \) is that the average speed of drivers is greater than 25 mph, so \( H_a: \mu > 25 \).
02

Calculate the Sample Mean

Calculate the mean of the given speeds. \[ \text{Mean} = \frac{27 + 33 + 32 + 21 + 30 + 30 + 29 + 25 + 27 + 34}{10} = 28.8 \text{ mph} \].
03

Calculate the Standard Deviation and Standard Error

First, calculate the standard deviation (\( s \)) of the sample. Then calculate the standard error (SE) using \( SE = \frac{s}{\sqrt{n}} \), where \( n = 10 \). Use the formula for standard deviation: \[ s = \sqrt{\frac{\sum (x_i - \text{mean})^2}{n-1}} \].For this dataset, \( s \approx 4.08 \). Thus, \[ SE = \frac{4.08}{\sqrt{10}} \approx 1.29 \].
04

Calculate the Test Statistic

Use the formula for the t-statistic: \[ t = \frac{\text{sample mean} - \text{hypothesized mean}}{SE} = \frac{28.8 - 25}{1.29} \approx 2.95 \].
05

Find the P-value

Refer to the t-distribution table with \( n-1 = 9 \) degrees of freedom for \( t = 2.95 \). This value yields a p-value less than 0.01.
06

Make a Decision

If the p-value is less than the significance level \( \alpha = 0.05 \), we reject the null hypothesis. Since the p-value is less than 0.01, we reject \( H_0 \) and conclude there is convincing evidence that the average speed is greater than 25 mph.
07

Evaluate Potential Errors

Since we rejected the null hypothesis, a Type I error could have been made. A Type I error occurs when we incorrectly reject a true null hypothesis. In context, this means that we concluded the driver's average speed is greater than 25 mph when it actually is not.

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Key Concepts

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

Type I and Type II errors
When conducting hypothesis testing, understanding errors is crucial. There are two primary types: Type I errors and Type II errors.
- **Type I Error**: This occurs when we incorrectly reject a true null hypothesis. In this specific context, a Type I error would mean concluding that the average speed in the construction zone is greater than 25 mph, when in fact, it is not. The consequences could involve unnecessary changes in enforcement or policies based on incorrect assumptions.
- **Type II Error**: This happens when we fail to reject a false null hypothesis. In the construction zone example, a Type II error would mean not concluding that the average speed is greater than 25 mph when it actually is. This could lead to inadequate safety measures being applied.
Understanding these errors helps in interpreting the results of statistical tests more effectively and deciding on appropriate actions.
Statistical Significance
Statistical significance helps us understand whether our results can be considered substantial or occurred by random chance. A result is statistically significant when the p-value is less than a pre-determined significance level, often denoted as \( \alpha \).
In this exercise, the significance level is set at 0.05. With a p-value less than 0.01, the result is statistically significant. This indicates strong evidence against the null hypothesis that the average speed is 25 mph. We can be confident that the observed average speed of 28.8 mph is likely greater than 25 mph and not due to random chance.
Statistical significance provides the backbone for scientific evidence, allowing researchers and analysts to make informed conclusions based on data.
t-test
A t-test is used to determine if there is a significant difference between the means of two groups, or a group's mean and a known value. Here, we're using a one-sample t-test to check if the sample mean differs significantly from the hypothesized population mean.
- **Test Statistic Calculation**: The t-statistic is calculated as \( t = \frac{\text{sample mean} - \text{hypothesized mean}}{SE} \). In our example, the t-statistic is approximately 2.95.
- **Degrees of Freedom**: It's important to consider degrees of freedom, which for a one-sample t-test is \( n-1 \), where \( n \) is the sample size. Here, it is 9.
The t-test helps determine whether the observed differences are statistically significant, considering sample variability and size.
Standard Deviation
Standard deviation quantifies the amount of variation or dispersion in a set of data values. A low standard deviation indicates that data points tend to be close to the mean, while a high standard deviation indicates a wider spread.
To calculate standard deviation, you use the formula \( s = \sqrt{\frac{\sum (x_i - \text{mean})^2}{n-1}} \). For our speeds, \( s \approx 4.08 \).
- **Standard Error (SE)**: This is the standard deviation of the sample mean's distribution. It's calculated as \( SE = \frac{s}{\sqrt{n}} \). SE plays a crucial role in hypothesis testing and helps determine the precision of the sample mean estimate.
Understanding standard deviation enhances our comprehension of data variability and the reliability of statistical conclusions.

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Most popular questions from this chapter

For the study of Jordanian children in Exercise 2 , the sample mean hemoglobin level was \(11.3 \mathrm{~g} / \mathrm{dl}\) and the sample standard deviation was \(1.6 \mathrm{~g} / \mathrm{dl} .\) A significance test yields a \(P\) -value of 0.0016 . (a) Explain what it would mean for the null hypothesis to be true in this setting. (b) Interpret the \(P\) -value in context.

A drug manufacturer claims that fewer than \(10 \%\) of patients who take its new drug for treating Alzheimer's disease will experience nausea. To test this claim, a significance test is carried out of $$ \begin{array}{l} H_{0}: p=0.10 \\ H_{a}: p<0.10 \end{array} $$ You learn that the power of this test at the \(5 \%\) significance level against the alternative \(p=0.08\) is 0.29 . (a) Explain in simple language what "power \(=0.29 "\) means in this setting. (b) You could get higher power against the same alternative with the same \(\alpha\) by changing the number of measurements you make. Should you make more measurements or fewer to increase power? Explain. (c) If you decide to use \(\alpha=0.01\) in place of \(\alpha=0.05\), with no other changes in the test, will the power increase or decrease? Justify your answer. (d) If you shift your interest to the alternative \(p=0.07\) with no other changes, will the power increase or decrease? Justify your answer.

The French naturalist Count Buffon \((1707-1788)\) tossed a coin 4040 times. He got 2048 heads. That's a bit more than one-half. Is this evidence that Count Buffon's coin was not balanced? To find out, Luisa decides to perform a significance test. Unfortunately, she made a few errors along the way. Your job is to spot the mistakes and correct them. $$ \begin{array}{l} H_{0}: \mu>0.5 \\ H_{a}: \bar{x}=0.5 \end{array} $$ \(\bullet\quad\) \(10 \%: 4040(0.5)=2020\) and \(4040(1-0.5)=2020\) are both at least 10 . \(\bullet\quad\) Large Counts: There are at least 40,400 coins in the world. \(t=\frac{0.5-0.507}{\sqrt{\frac{0.5(0.5)}{4040}}}=-0.89 ; P\) -value \(=1-0.1867=0.8133\) Reject \(H_{0}\) because the \(P\) -value is so large and conclude that the coin is fair.

Walking to school A recent report claimed that \(13 \%\) of students typically walk to school. \({ }^{10}\) DeAnna thinks that the proportion is higher than 0.13 at her large elementary school, so she surveys a random sample of 100 students to find out.

The \(z\) statistic for a test of \(H_{0}: p=0.4\) versus \(H_{a}: p \neq 0.4\) is \(z=2.43 .\) This test is (a) not significant at either \(\alpha=0.05\) or \(\alpha=0.01\). (b) significant at \(\alpha=0.05\) but not at \(\alpha=0.01\). (c) significant at \(\alpha=0.01\) but not at \(\alpha=0.05\). (d) significant at both \(\alpha=0.05\) and \(\alpha=0.01\). (e) inconclusive because we don't know the value of \(\hat{p}\).

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