/*! 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 24 A motorist claims that the South... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A motorist claims that the South Boro Police issue an average of 60 speeding tickets per day. These data show the number of speeding tickets issued each day for a randomly selected period of 30 days. Assume \(\sigma\) is \(13.42 .\) Is there enough evidence to reject the motorist's claim at \(\alpha=0.05 ?\) Use the \(P\) -value method. \(\begin{array}{llllllll}72 & 45 & 36 & 68 & 69 & 71 & 57 & 60 \\ 83 & 26 & 60 & 72 & 58 & 87 & 48 & 59 \\ 60 & 56 & 64 & 68 & 42 & 57 & 57 & \\ 58 & 63 & 49 & 73 & 75 & 42 & 63 & \end{array}\)

Short Answer

Expert verified
There is not enough evidence to reject the motorist's claim.

Step by step solution

01

Formulate the Hypotheses

First, we need to establish the null and alternative hypotheses. The null hypothesis \( H_0 \) states that the average number of tickets issued per day is equal to the motorist's claim, which is 60 tickets per day (\( \mu = 60 \)). The alternative hypothesis \( H_a \) states that the average number of tickets per day is not equal to 60 (\( \mu eq 60 \)).
02

Gather Information and Calculate the Sample Mean

Collect the data from the problem: \( n = 30 \) (number of days), \( \sigma = 13.42 \), and the sample values. Compute the sample mean \( \bar{x} \). The sample mean \( \bar{x} \) is calculated as \[ \bar{x} = \frac{\sum x_i}{n} = \frac{72 + 45 + 36 + \ldots + 63}{30} = 60.33 \] after summing the values 1810 and dividing by 30.
03

Calculate the Standard Error and Test Statistic

The standard error of the mean (SE) is calculated using the formula \[ SE = \frac{\sigma}{\sqrt{n}} = \frac{13.42}{\sqrt{30}} \approx 2.45 \]. Now calculate the test statistic \( z \) using the formula \[ z = \frac{\bar{x} - \mu}{SE} = \frac{60.33 - 60}{2.45} \approx 0.13 \].
04

Find the P-value

Using the calculated test statistic \( z = 0.13 \) and a two-tailed test (since \( \mu eq 60 \)), find the \( P \)-value. The \( P \)-value for \( z = 0.13 \) is approximately 0.8962, using a standard normal distribution table or a calculator.
05

Compare the P-value with Alpha

Now compare the \( P \)-value with the significance level \( \alpha = 0.05 \). Since \( P = 0.8962 \) is greater than \( \alpha = 0.05 \), we fail to reject the null hypothesis.

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.

Null Hypothesis
In hypothesis testing, we start with what we call the null hypothesis, denoted as \( H_0 \). This is our starting assumption that there is no effect or no difference – essentially a status quo claim. For the problem about speeding tickets, the null hypothesis asserts that the average number of tickets issued each day is as the motorist claims, 60. Formally, it is written as \( \mu = 60 \), where \( \mu \) is the population mean number of tickets.
The null hypothesis is crucial because it sets the stage for our testing. We assume it's true, and then use statistical evidence to determine if there is enough reason to doubt it. If the evidence suggests otherwise, we might reject the null hypothesis, favoring the alternative.
Alternative Hypothesis
The alternative hypothesis, denoted as \( H_a \), challenges the claim made by the null hypothesis. It proposes that the actual population parameter does indeed differ from the stated value. In our exercise about speeding tickets, the alternative hypothesis suggests that the average daily number of tickets is not as the motorist claims. We express it as \( \mu eq 60 \).
The purpose of the alternative hypothesis is to establish a competing claim. If the statistical evidence is strong enough, it leads us to reject the null hypothesis in favor of this alternative. It's essential as it represents the conclusion to be tested or the statement expected to be supported by evidence.
P-value Method
The \( P \)-value method is a fundamental tool in hypothesis testing. It helps us determine the significance of our test results. The \( P \)-value is the probability that the observed data (or something more extreme) would occur if the null hypothesis were true.
  • If the \( P \)-value is less than the significance level \( \alpha \) (in this case, 0.05), we conclude there's enough evidence to reject the null hypothesis.
  • If it's higher, as we found with a \( P \)-value of 0.8962, we fail to reject the null hypothesis, suggesting that there isn't sufficient evidence to dispute the motorist's claim.
Using the \( P \)-value method offers a clear criterion for decision-making that relies on probability, making hypothesis tests clear and interpretable.
Standard Error
Standard error is a measure of the variability or dispersion of a sample mean estimate around the population mean. It gives an indication of how much the sample mean might differ from the true population mean. The formula for standard error when the population standard deviation \( \sigma \) is known is:
\[ SE = \frac{\sigma}{\sqrt{n}} \]where \( n \) is the sample size.
In our exercise, the standard error calculated is approximately 2.45. This tells us about the margin of error for our sample mean estimate from the true population mean. It's a crucial component in calculating the test statistic \( z \), which helps us find the \( P \)-value for hypothesis testing. Understanding standard error aids in appreciating the precision of our sample estimates.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Perform each of the following steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Researchers suspect that \(18 \%\) of all high school students smoke at least one pack of cigarettes a day. At Wilson High School, a randomly selected sample of 300 students found that 50 students smoked at least one pack of cigarettes a day. At \(\alpha=0.05,\) test the claim that less than \(18 \%\) of all high school students smoke at least one pack of cigarettes a day. Use the \(P\) -value method.

Define null and alternative hypotheses, and give an example of each.

In hypothesis testing, why can't the hypothesis be proved true?

Perform each of the following steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Find the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Assume that the population is approximately normally distributed. The Old Farmer's Almanac stated that the average consumption of water per person per day was 123 gallons. To test the hypothesis that this figure may no longer be true, a researcher randomly selected 16 people and found that they used on average 119 gallons per day and \(s=5.3 .\) At \(\alpha=0.05,\) is there enough evidence to say that the Old Farmer's Almanac figure might no longer be correct? Use the \(P\) -value method.

The average expenditure per student (based on average daily attendance) for a certain school year was \(\$ 10,337\) with a population standard deviation of \(\$ 1560 .\) A survey for the next school year of 150 randomly selected students resulted in a sample mean of \(\$ 10,798\). Do these results indicate that the average expenditure has changed? Choose your own level of significance.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.