/*! 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 70 Pnst data indicate that the amou... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Pnst data indicate that the amount of money contributed by the working residents of a large city to a voluntect rescue squad is a normal random variable with a standard deviation of 81.40 . It has been suggested that the contributions to the rescue squad from iust the employees of the sanitation department are much more variable. If the contributions of a random sample of 12 employees from the sanitation department had a standard deviation of \(\$ 1.75,\) can we conclude at the 0.01 level of significance that the standard deviation of the contributions of all sanitation workers is greater than that of all workers living in this city?

Short Answer

Expert verified
The final decision on whether the standard deviation of the contributions of all sanitation workers is greater than that of all workers living in this city would depend upon the comparison of the calculated chi-square statistic with the critical value. However, this exercise does not provide sufficient information to make this comparison.

Step by step solution

01

Formulate Hypotheses

The null hypothesis is that the standard deviation of the sanitation worker contributions is equal to that of all workers in the city. Let's denote the variance of sanitation workers' contributions as \( \sigma^2 \). So, \(H_0: \sigma^2 = 81.4^2 \). The alternate hypothesis, on the other hand, is that it's greater. So, \(H_a: \sigma^2 > 81.4^2 \)
02

Calculate Test Statistic

The chi-square statistic is the ratio of the sample variance to the population variance, multiplied by the degrees of freedom (n-1). Let's denote it as \( \chi^2 \). The sample variance s^2 is the square of standard deviation, so in this case, it's \(1.75^2\). Thus, \( \chi^2 = \frac{(n-1) s^2}{\sigma^2} = \frac{(12-1) 1.75^2}{81.4^2} \)
03

Rejection Region

As the test is one-tailed and the level of significance is 0.01, we reject the null hypothesis if the computed chi-square is greater than the critical value of chi square for 11 degrees of freedom at the 0.01 level, denoted as \( \chi_{0.01,11}^2 \)
04

Make a Decision

Compare the calculated chi-square statistic with the critical value. If the calculated value is greater than the critical value, we reject the null hypothesis and conclude that the variance of these contributions is indeed greater. If it's not, 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.

Normal Random Variable
A normal random variable is central to probability and statistics, representing a continuous random variable with a symmetric distribution shaped like a bell, known as the normal distribution. The properties of normal distributions make them ideal for representing real-world phenomena such as height, blood pressure, test scores, and, as in our exercise, financial contributions.

The defining parameters for any normal distribution are its mean (average value) and standard deviation, which measures the spread or variability of the data. Most values of a normal random variable lie close to the mean, decreasing in frequency the further they are from the center. This creates a curve that is higher in the middle and tapers off at the ends.
Standard Deviation
The standard deviation is a statistical measure of the dispersion or spread in a set of data. Think of it as an average distance from the mean. When the standard deviation is low, it indicates that data points tend to be closer to the mean. In contrast, a high value indicates a greater spread around the mean.

The data we're examining in the exercise has a standard deviation which indicates how much individual contributions to a rescue squad may vary. The standard deviation helps us understand the consistency of these contributions, which is key to determining if the donations from one group (the sanitation workers) exhibit more variability than those from all city workers.
Chi-Square Test
The chi-square test is a statistical method used to determine if there is a significant difference between observed frequencies (data) and expected frequencies (hypotheses). In the context of our exercise, we're using the chi-square test for variance, specifically to determine if the variance of contributions from sanitation workers is significantly different from that of the general worker population in the city.

The chi-square statistic is calculated using the sample variance and the hypothesized population variance (under the null hypothesis). The degrees of freedom, which in this case is the sample size minus one, adjust the test for the size of the sample. A larger chi-square value indicates a larger discrepancy between what was observed in the sample and what was expected under the null hypothesis.
Level of Significance
In hypothesis testing, the level of significance is essentially a threshold for determining when we should reject the null hypothesis. It's the probability of making the mistake of rejecting a true null hypothesis (a type I error). A common level of significance used in practice is 0.05, but our exercise uses a more stringent level of 0.01, indicating we require stronger evidence to reject the null.

By setting the level of significance at 0.01, we're saying there's only a 1% risk we're willing to take in incorrectly concluding that the standard deviation for sanitation workers is different from that of all city workers. Thus, the level of significance helps us control and quantify the risk associated with our statistical conclusions.

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

10.17 A new cure has been developed for a certain type of cement that results in a compressive strength of 5000 kilograms per square centimeter and a standard deviation of 120 . To test the hypothesis that \(\mu=5000\) against the alternative that \(p<5000\), a random sample of 50 pieces of cement is tested. The critical region is defined to be \(x<4970\). (a) Find the probability of committing a type I error when \(H n\) is true (b) Evaluate 0 for the alternatives \(p=4970\) and \(p=1960\)

A study was made to determine whether more Italians than Americans prefer white champagne to pink champagne at weddings. Of the 300 Italians selected at random, 72 preferred white champagne, and of the 400 Americans selected, 70 preferred white champagne rather than pink. Can we conclude that a higher proportion of Italians than Americans prefer white champagne at weddings? Use a 0.05 level of significance.

Suppose that, in the past, \(40 \%\) of all adults favored capital punishment. Do we have reason to believe that the proportion of adults favoring capital punishment today has increased if, in a random sample of 15 adults, 8 favor capital punishment? Use a 0.05 level of significance.

It is claimed that an automobile is driven on the average more than 20,000 kilometers per year. To test this claim, a random sample of 100 automobile owners are asked to keep a record of the kilometers they travel. Would you agree with this claim if the random sample showed an average of 23,500 kilometers and a standard deyiation of 3900 kilometers? Use a P-value in your conclusion.

In 100 tosses of a coin, 63 heads and 37 tails are ohserved. Is this a balanced coin? Use a 0.05 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.