/*! 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 2 A simple random sample of size \... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A simple random sample of size \(n=200\) individuals with a valid driver's license is asked if they drive an American-made automobile. Of the 200 individuals surveyed, 115 responded that they drive an American-made automobile. Determine if a majority of those with a valid driver's license drive an American made automobile at the \(\alpha=0.05\) level of significance.

Short Answer

Expert verified
Yes, a majority of individuals with a valid driver's license drive an American-made automobile at the 0.05 level of significance.

Step by step solution

01

State the hypotheses

Formulate the null and alternative hypotheses. The null hypothesis states that 50% or fewer of the surveyed individuals have an American-made automobile, while the alternative hypothesis states that more than 50% of the surveyed individuals have an American-made automobile.Null hypothesis: \[ H_0: p \leq 0.5 \]Alternative hypothesis: \[ H_1: p > 0.5 \]
02

Calculate the sample proportion

Compute the sample proportion (\( \hat{p} \)), which is the number of individuals who drive an American-made automobile divided by the sample size.\[ \hat{p} = \frac{115}{200} = 0.575 \]
03

Calculate the standard error

The standard error (\( \text{SE} \)) of the sample proportion is given by \( \text{SE} = \sqrt{\frac{p \(1 - p\)}{n}} \), assuming \( p = 0.5 \) under the null hypothesis.\[ \text{SE} = \sqrt{\frac{0.5 \(1 - 0.5\)}{200}} = 0.0354 \]
04

Compute the test statistic

The test statistic (\( z \)) can be found using the formula \( z = \frac{\hat{p} - p}{\text{SE}} \).\[ z = \frac{0.575 - 0.5}{0.0354} = 2.12 \]
05

Find the critical value

For a one-tailed test at the \( \alpha = 0.05 \) significance level, the critical value (\( z_{\alpha} \)) is 1.645.
06

Compare the test statistic to the critical value

Compare the computed test statistic to the critical value. If the test statistic is greater than the critical value, reject the null hypothesis. Since \( z=2.12 \) is greater than \( z_{\alpha} = 1.645 \), we reject the null hypothesis.
07

State the conclusion

Since we reject the null hypothesis, we conclude that there is sufficient evidence at the \( \alpha = 0.05 \) level of significance to support the claim that a majority of individuals with a valid driver's license drive an American-made automobile.

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.

Simple Random Sample
In hypothesis testing, it's crucial to have a sample that is representative of the population. A simple random sample ensures this by selecting individuals in such a way that every individual in the population has an equal chance of being chosen. This helps in reducing bias.

For example, in our exercise, a sample of 200 individuals with a valid driver's license was selected randomly. This randomness ensures that the sample is likely to reflect the true characteristics of the entire population of licensed drivers.
Proportion
A proportion in statistics refers to the fraction of the total that has a particular attribute. It's a way to express part of a whole in percentage terms.

In the exercise, the proportion of individuals who drive an American-made automobile was calculated using the formula:
\( \hat{p} = \frac{115}{200} = 0.575 \)

This means that 57.5% of the sampled individuals reported driving an American-made automobile.
Standard Error
The standard error (\text{SE}) measures the accuracy with which a sample proportion represents the true population proportion. It takes into account both the variability within the population and the sample size.

To calculate the standard error, we use the formula:
\[ \text{SE} = \sqrt{\frac{p (1 - p)}{n}} \] Assuming \( p = 0.5 \) under the null hypothesis, we get:
\[ \text{SE} = \sqrt{\frac{0.5 (1 - 0.5)}{200}} = 0.0354 \]

This provides a measure of how much we expect our sample proportion to fluctuate if we were to take many samples.
Test Statistic
The test statistic is a standardized value that is calculated from sample data during a hypothesis test. It measures how far the sample statistic is from the null hypothesis value, in terms of the standard error.

In the exercise, the test statistic (\text{z}) is calculated as:
\[ \text{z} = \frac{\hat{p} - p}{\text{SE}} \]
Substituting the values, we get:
\[ \text{z} = \frac{0.575 - 0.5}{0.0354} = 2.12 \]

This value tells us how many standard errors the sample proportion is away from the hypothesized population proportion.
Level of Significance
The level of significance (\text{α}) is a threshold set by the researcher before conducting a hypothesis test. It indicates the probability of rejecting the null hypothesis when it is actually true, i.e., making a Type I error.

In the exercise, a significance level of \( \alpha = 0.05 \) was used. This corresponds to a 5% risk of concluding that a majority of licensed drivers prefer American-made cars when they actually do not.

To make our decision, we compare the test statistic to the critical value associated with \( \alpha = 0.05 \). If the test statistic exceeds this critical value, we reject the null hypothesis. Here, because \( \text{z} = 2.12 \) is greater than the critical value of 1.645, we reject the null hypothesis, concluding that a majority do prefer American-made automobiles.

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

Suppose an acquaintance claims to have the ability to determine the birth month of randomly selected individuals. To test such a claim, you randomly select 80 individuals and ask the acquaintance to state the birth month of the individual. If the individual has the ability to determine birth month, then the proportion of correct birth months should exceed \(\frac{1}{12},\) the rate one would expect from simply guessing. (a) State the null and alternative hypotheses for this experiment. (b) Suppose the individual was able to guess nine correct birth months. The \(P\) -value for such results is \(0.1726 .\) Explain what this \(P\) -value means and write a conclusion for the test.

Explain the difference between statistical significance and practical significance.

Test the hypothesis using (a) the classical approach and (b) the P-value approach. Be sure to verify the requirements of the test. $$\begin{array}{l}H_{0}: p=0.55 \text { versus } H_{1}: p<0.55 \\\n=150 ; x=78 ; \alpha=0.1\end{array} $$

In Problems \(9-14,\) the null and alternative hypotheses are given. Determine whether the hypothesis test is lefi-tailed, right-tailed, or two-tailed. What parameter is being tested? \(H_{0}: \sigma=7.8\) \(H_{1}: \sigma \neq 7.8\)

ACT, a college entrance exam used for admission, looked at historical records and established 21 as the minimum score on the ACT reading portion of the exam for a student to be considered prepared for social science in college. (Note: "Being prepared" means there is a \(75 \%\) probability of successfully completing a social science course in college.) An official with the Illinois State Department of Education wonders whether a majority of the students in her state who took the \(\mathrm{ACT}\) are prepared to take social science. She obtains a simple random sample of 500 records of students who have taken the \(\mathrm{ACT}\) and finds that 269 are prepared. Does this represent significant evidence that a majority (more than \(50 \%\) ) of the students in the state of Illinois are prepared for social science in college upon graduation? (a) What does it mean to make a Type II error for this test? (b) If the researcher decides to test this hypothesis at the \(\alpha=0.05\) level of significance, determine the probability of making a Type II error if the true population proportion is 0.52. What is the power of the test? (c) Redo part (b) if the true proportion is \(0.55 .\)

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.