/*! 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 50 The policy of the Suburban Trans... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 50

The policy of the Suburban Transit Authority is to add a bus route if more than 55 percent of the potential commuters indicate they would use the particular route. A sample of 70 commuters revealed that 42 would use a proposed route from Bowman Park to the downtown area. Does the Bowman-to- downtown route meet the STA criterion? Use the .05 significance level.

Short Answer

Expert verified
Yes, the Bowman-to-downtown route meets the STA criterion, as the sample shows more than 55% would use it and the test statistic exceeds the critical value.

Step by step solution

01

Define the Hypotheses

First, define the null and alternative hypotheses. The null hypothesis, denoted as \( H_0 \), states that the proportion of commuters who would use the new bus route is 55% or less, i.e., \( p \leq 0.55 \). The alternative hypothesis, denoted as \( H_a \), states that the proportion is greater than 55%, i.e., \( p > 0.55 \).
02

Determine the Sample Proportion

Compute the sample proportion \( \hat{p} \) using the formula \( \hat{p} = \frac{x}{n} \), where \( x = 42 \) (the number of commuters who would use the route) and \( n = 70 \) (the total number of surveyed commuters). Thus, \( \hat{p} = \frac{42}{70} = 0.6 \).
03

Calculate the Test Statistic

Use the formula for the test statistic in a one-sample proportion test: \( z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \), where \( p_0 = 0.55 \). Substitute to find \( z = \frac{0.6 - 0.55}{\sqrt{\frac{0.55(0.45)}{70}}} \). Calculate to obtain the z-value.
04

Find the Critical Value

Determine the critical value for a one-tailed test at the 0.05 significance level. A common z-table or statistical software can be used to find that the critical z-value (for z > critical value in a right-tailed test) is approximately 1.645.
05

Compare Test Statistic and Critical Value

Compare the computed z-statistic to the critical value from the z-table. If the z-statistic is greater than the critical z-value, reject the null hypothesis.
06

Conclusion

Based on the comparison, if the z-value calculated is greater than 1.645, the null hypothesis is rejected, meaning that enough evidence exists to support the STA's criterion that more than 55% will use the route. Otherwise, it fails to meet the criterion.

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 91影视!

Key Concepts

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

Significance Level
In hypothesis testing, the significance level, often denoted as \( \alpha \), is crucial as it helps determine the threshold for deciding if an observed effect is statistically significant. For this exercise, we are using a 0.05 significance level.

Here's what it means:
  • If the probability of observing our data, assuming the null hypothesis is true, is less than 5%, we reject the null hypothesis.
  • This level of significance is a balance between Type I errors (false positives) and Type II errors (false negatives).
  • In more practical terms, at a 0.05 level, there's a 5% risk of concluding that there is an effect when there is actually none.
Choosing the right significance level is like setting a standard. A 0.05 threshold is commonly used in many fields, including public transit policies, as in this exercise.
Proportion Test
A proportion test evaluates hypotheses about a population proportion. It helps to determine if the observed proportion significantly differs from a known or assumed value.

When performing a proportion test, follow these general steps:
  • First, identify the sample proportion \( \hat{p} \). In this exercise, it was calculated as \( \hat{p} = \frac{42}{70} = 0.6 \).
  • Compare this sample proportion to the population proportion \( p_0 \), which is hypothesized (in this case, 0.55).
This test is crucial because it uses the information from a sample to make inferences about a larger population. It's like zooming out from a small picture to make conclusions about the bigger image. In this context, you determine if more than 55% of commuters would use the bus route based on the sample.
Null and Alternative Hypotheses
The null and alternative hypotheses are the foundation of hypothesis testing. They frame what you're looking to test. Each hypothesis acts as a possible answer to your question.

Null Hypothesis (\( H_0 \)):
  • Assumes no effect or change. In our exercise: the proportion (\( p \)) of commuters willing to use the bus route is 55% or less, i.e., \( p \leq 0.55 \).
Alternative Hypothesis (\( H_a \)):
  • Poses that there is an effect or a change. Here, it presumes that the proportion is greater than 55%, i.e., \( p > 0.55 \).
These hypotheses are mutually exclusive. If the evidence from data is strong enough, you will reject the null hypothesis and accept the alternative. This helps in decision-making, like determining whether to establish a new bus route based on commuter interest.
Test Statistic Calculation
Calculating the test statistic is a critical step in determining whether the sample data provides enough evidence to reject the null hypothesis.

In the case of a proportion test, the test statistic (\( z \)) formula is:
  • \( z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \)
Where:
  • \( \hat{p} \) is the sample proportion, calculated here as \( 0.6 \).
  • \( p_0 \) is the hypothesized population proportion value (0.55).
  • \( n \) is the sample size (70 in this exercise).
Plugging in the values gives you the test statistic. Compare this \( z \)-value to the critical \( z \)-value from a standard normal distribution to make your final hypothesis decision.

The test statistic acts like a measuring stick. It helps you see how far your sample proportion is from the hypothesized proportion, in units of standard error. This aids in deciding whether the observed data is typical under the null hypothesis or not.

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

A spark plug manufacturer claimed that its plugs have a mean life in excess of 22,100 miles. Assume the life of the spark plugs follows the normal distribution. A fleet owner purchased a large number of sets. A sample of 18 sets revealed that the mean life was 23,400 miles and the standard deviation was 1,500 miles. Is there enough evidence to substantiate the manufacturer's claim at the .05 significance level?

Experience raising New Jersey Red chickens revealed the mean weight of the chickens at five months is 4.35 pounds. The weights follow the normal distribution. In an effort to increase their weight, a special additive is added to the chicken feed. The subsequent weights of \(a\) sample of five-month- old chickens were (in pounds): $$\begin{array}{|lllllllll|}\hline 4.41 & 4.37 & 4.33 & 4.35 & 4.30 & 4.39 & 4.36 & 4.38 & 4.40 & 4.39 \\\\\hline\end{array}$$ At the .01 level, has the special additive increased the mean weight of the chickens? Estimate the \(p\) -value.

The liquid chlorine added to swimming pools to combat algae has a relatively short shelf life before it loses its effectiveness. Records indicate that the mean shelf life of a 5 -gallon jug of chlorine is 2,160 hours (90 days). As an experiment, Holdlonger was added to the chlorine to find whether it would increase the shelf life. A sample of nine jugs of chlorine had these shelf lives (in hours): $$\begin{array}{|lllllllll|}\hline 2,159 & 2,170 & 2,180 & 2,179 & 2,160 & 2,167 & 2,171 & 2,181 & 2,185 \\\\\hline\end{array}$$ At the .025 level, has Holdlonger increased the shelf life of the chlorine? Estimate the \(p\) -value.

A statewide real estate sales agency, Farm Associates, specializes in selling farm property in the state of Nebraska. Its records indicate that the mean selling time of farm property is 90 days. Because of recent drought conditions, the agency believes that the mean selling time is now greater than 90 days. A statewide survey of 100 farms sold recently revealed that the mean selling time was 94 days, with a standard deviation of 22 days. At the . 10 significance level, has there been an increase in selling time?

Given the following hypothesis: $$\begin{array}{l}H_{i} ; \mu=100 \\\H_{1}: \mu \neq 100\end{array}$$ A random sample of six resulted in the following values: 118 \(105,112,119,105,\) and \(111 .\) Using the .05 significance level, can we conclude the mean is different from \(100 ?\) a. State the decision rule. b. Compute the value of the test statistic. c. What is your decision regarding the null hypothesis? d. Estimate the \(p\) -value.
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Logic and Functions

Read Explanation

Pure Maths

Read Explanation

Calculus

Read Explanation

Statistics

Read Explanation
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.