/*! 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 15 For Exercises 7 through \(27,\) ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

For Exercises 7 through \(27,\) perform these 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. Commuters A recent random survey of 100 individuals in Michigan found that 80 drove to work alone. A similar survey of 120 commuters in New York found that 62 drivers drove alone to work. Find the \(95 \%\) confidence interval for the difference in proportions.

Short Answer

Expert verified
There is a significant difference between the proportions of commuters driving alone in Michigan and New York.

Step by step solution

01

State the hypotheses

Here, we are comparing two proportions. Let \( p_1 \) be the proportion of individuals in Michigan who drive alone, and \( p_2 \) be the proportion of individuals in New York who drive alone. The null hypothesis is \( H_0: p_1 = p_2 \), and the alternative hypothesis is \( H_1: p_1 eq p_2 \). We are interested in a two-tailed test as we are checking for any difference.
02

Identify the claim

The claim in this scenario is that there is a difference between the proportions of individuals who drive to work alone in Michigan and New York, hence \( H_1: p_1 eq p_2 \).
03

Find the critical value(s)

For a two-tailed test with a significance level of \( \alpha = 0.05 \), the critical values for \( z \) are \( \pm 1.96 \), which corresponds to the 95% confidence level.
04

Compute the test value

First, calculate the sample proportions: \( \hat{p}_1 = \frac{80}{100} = 0.8 \) and \( \hat{p}_2 = \frac{62}{120} = 0.5167 \). The pooled proportion \( \hat{p} \) is calculated as \( \hat{p} = \frac{80 + 62}{100 + 120} = \frac{142}{220} \approx 0.645 \). Now, compute the test statistic using the formula \[ z = \frac{{\hat{p}_1 - \hat{p}_2}}{\sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}\]Substitute the values: \[ z = \frac{0.8 - 0.5167}{\sqrt{0.645 \times 0.355 \left(\frac{1}{100} + \frac{1}{120}\right)}} \approx \frac{0.2833}{0.0675} \approx 4.20\]
05

Make the decision

Compare the calculated test statistic \( z = 4.20 \) with the critical values \( -1.96 \) and \( 1.96 \). Since \( 4.20 \) is greater than \( 1.96 \), we reject the null hypothesis \( H_0 \).
06

Summarize the results

At the 95% confidence level, there is enough evidence to reject the null hypothesis and conclude that there is a significant difference in the proportions of individuals driving to work alone in Michigan and New York.

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.

Confidence Intervals
Confidence intervals are an important concept in statistics. They offer a range of values which estimate an unknown parameter, such as a population proportion. This range expresses where the true value likely lies, considering a certain level of confidence (like 95%).
A 95% confidence interval indicates that if we took many samples and calculated intervals for each one, about 95% of those intervals would contain the true population parameter. Confidence intervals provide more information than a simple point estimate by incorporating sampling variability.
In the case of Michigan vs. New York drivers, the 95% confidence interval helps quantify the difference in the proportion of those driving alone. This interval helps us determine whether the observed difference is statistically significant or might have occurred by random chance.
Two-Tailed Test
A two-tailed test is used in hypothesis testing to determine if there is a significant difference in either direction from a specified value. When applying a two-tailed test, researchers are concerned with deviations in both directions - above or below the hypothesized parameter.
The test derives its name from the fact that the critical area, which dictates the rejection of the null hypothesis, is located in both tails of the distribution curve. This ensures if there is an extreme sample outcome at either tail, the result will be significant.
In the exercise, where the proportions of solo drivers in two states are compared, a two-tailed test checks for significant differences either way: whether Michigan has a higher or lower proportion of drivers compared to New York. This broadens the scope, making the test applicable to any shifts from the hypothesized equality.
Proportions
Proportions represent parts or fractions of a whole, often expressed as percentages. In statistics, proportion comparisons often deal with data concerning populations or samples.
Consider the exercise - where the proportion of people who commute alone is measured for Michigan and New York. Here, we're focused on the general proportion formula: \[ \hat{p} = \frac{\text{Number of successes}}{\text{Sample size}} \]For Michigan, \(\hat{p}_1 = \frac{80}{100} = 0.8 \), while for New York, \(\hat{p}_2 = \frac{62}{120} \approx 0.5167 \).
This calculation provides essential information about each group's behavior, vital in comparing differences between two groups in hypothesis testing.
Significance Level
The significance level, denoted by \( \alpha \), is a threshold for determining the cutoff point at which the null hypothesis can be rejected. Often set at 0.05, it indicates a 5% risk of rejecting a true null hypothesis.
The chosen significance level dictates the boundaries of the critical region for a hypothesis test. A lower \( \alpha \) means more strict criteria for rejecting a null hypothesis, thus a decreased chance of Type I errors (false positives).
In our exercise, a significance level of 0.05 was used. This resulted in critical values of \( \pm 1.96 \) for the z-distribution. It helped in making a decision, since the calculated test statistic \( z = 4.20 \) was greater than \( 1.96 \).
This demonstrated that, given the selection of \( \alpha \), it was highly unlikely the observed differences were purely due to random chance.

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

For Exercises 2 through \(12,\) perform each of these steps. Assume that all variables are normally or approximately normally distributed. 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. Mistakes in a Song A random sample of six music students played a short song, and the number of mistakes in music each student made was recorded. After they practiced the song 5 times, the number of mistakes each student made was recorded. The data are shown. At \(\alpha=0.05,\) can it be concluded that there was a decrease in the mean number of mistakes? $$ \begin{array}{l|cccccc}{\text { Student }} & {\mathrm{A}} & {\mathrm{B}} & {\mathrm{C}} & {\mathrm{D}} & {\mathrm{E}} & {\mathrm{F}} \\ \hline \text { Before } & {10} & {6} & {8} & {8} & {13} & {8} \\ \hline \text { After } & {4} & {2} & {2} & {7} & {8} & {9}\end{array} $$

Explain the difference between testing a single mean and testing the difference between two means.

Gasoline Prices A random sample of monthly gasoline prices was taken from 2011 and from 2015. The samples are shown. Using \(\alpha=0.01,\) can it be concluded that gasoline cost more in 2015? Use the \(P\) -value method. $$ \frac{2011}{2015} | \begin{array}{cccccc}{2011} & {2.02} & {2.47} & {2.50} & {2.70} & {3.13} & {2.56} \\ \hline 2015 & {2.36} & {2.46} & {2.63} & {2.76} & {3.00} & {2.85} & {2.77}\end{array} $$

What are the characteristics of the \(F\) distribution?

For Exercises 7 through \(27,\) perform these 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. Medical Supply Sales According to the U.S. Bureau of Labor Statistics, approximately equal numbers of men and women are engaged in sales and related occupations. Although that may be true for total numbers, perhaps the proportions differ by industry. A random sample of 200 salespersons from the industrial sector indicated that 114 were men, and in the medical supply sector, 80 of 200 were men. At the 0.05 level of significance, can we conclude that the proportion of men in industrial sales differs from the proportion of men in medical supply sales?

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.