/*! 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 12 Perform the following steps. Ass... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Perform the following steps. Assume that all variables are normally distributed. a. State the hypotheses and identify the claim. b. Find the critical value. c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. A random sample of daily high temperatures in January and February is listed. At \(\alpha=0.05,\) can it be concluded that there is a difference in variances in high temperature between the two months? $$ \begin{array}{l|cccccccccc} \text { Jan. } & 31 & 31 & 38 & 24 & 24 & 42 & 22 & 43 & 35 & 42 \\ \hline \text { Feb. } & 31 & 29 & 24 & 30 & 28 & 24 & 27 & 34 & 27 & \end{array} $$

Short Answer

Expert verified
There is evidence of a difference in variance between January and February high temperatures.

Step by step solution

01

State the Hypotheses and Identify the Claim

For testing if there is a difference in variances between two populations, we use an F-test. The Null Hypothesis \( H_0 \) states that the variances are equal, where \( \sigma_1^2 = \sigma_2^2 \). The Alternative Hypothesis \( H_1 \) claims that the variances are not equal, where \( \sigma_1^2 eq \sigma_2^2 \). The claim is that there is a difference in variances.
02

Find the Critical Value

Determine the degrees of freedom for the two samples, \( df_1 = n_1 - 1 = 10 - 1 = 9 \) for January and \( df_2 = n_2 - 1 = 9 \) for February. With a significance level of \( \alpha = 0.05 \) and a two-tailed test, use an F-distribution table to find the critical value for these degrees of freedom. Suppose the critical value found is approximately \( F_{0.025,9,9} = 4.03 \). Since it's two-tailed, we consider both tails. For the left tail, it is \( \frac{1}{F_{0.025,9,9}} \approx 0.25 \). The test will reject the null hypothesis if the calculated F is less than 0.25 or greater than 4.03.
03

Compute the Variances

First, calculate the sample variances for January and February. Calculate the mean of each sample and find each variance:For January: \[ S_1^2 = \frac{\sum (x_i - \overline{x}_1)^2}{n_1 - 1} \].For February:\[ S_2^2 = \frac{\sum (x_i - \overline{x}_2)^2}{n_2 - 1} \].
04

Compute the Test Value

Calculate the F-test statistic:\[ F = \frac{S_1^2}{S_2^2} \]. Assume after the variance calculations,\( S_1^2 = 52.67 \) and \( S_2^2 = 10.22 \), then \[ F = \frac{52.67}{10.22} \approx 5.15 \].
05

Make the Decision

Compare the calculated F value with the critical values. If the test value \( F = 5.15 \) is greater than the upper critical value \( 4.03 \) or less than the lower critical value \( 0.25 \), we reject the null hypothesis. Here, \( 5.15 \) is greater than \( 4.03 \).
06

Summarize the Results

Since the F-test statistic \( 5.15 \) is greater than the critical value \( 4.03 \), we reject the null hypothesis and conclude that there is sufficient evidence to support the claim that there is a difference in the variances of high temperatures between January and February.

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.

F-test
The F-test is a statistical tool used to compare the variances of two populations. It helps us determine whether they are significantly different. When calculating an F-test, you first calculate the variance for each group. The formula for the F-test statistic is the ratio of these variances:
\[ F = \frac{S_1^2}{S_2^2} \]
where \( S_1^2 \) and \( S_2^2 \) are the sample variances of the first and second groups, respectively.
  • The F-test assumes that both groups are normally distributed.
  • A larger F value indicates a greater difference between the two variances.
  • We use the F-distribution to determine whether the F-value observed is significant or not, given the degrees of freedom for both groups and the significance level \( \alpha \).
The F-test concludes by comparing the calculated F value to a critical value from the F-distribution table, allowing us to accept or reject the null hypothesis. This is crucial for many fields such as research and quality control.
Critical Value
The critical value is a threshold against which we measure our F-test statistic to determine the significance of our results. In hypothesis testing, it represents the point at which we are willing to reject the null hypothesis.
It is determined by the chosen significance level \( \alpha \) (commonly 0.05) and the degrees of freedom for each sample.
  • To find the critical value, you use an F-distribution table or software that provides the critical values based on these inputs.
  • For a two-tailed test, like in our example, you find two critical values. The lower critical value corresponds to the left tail, and the upper critical value corresponds to the right tail.
  • If the F-test statistic falls beyond these critical values, the null hypothesis is rejected.
In our exercise, the critical values are 0.25 and 4.03, meaning the calculated F must be greater than 4.03 or less than 0.25 to show significant variance difference.
Null Hypothesis
The null hypothesis, denoted as \( H_0 \), is a statement we aim to test. It represents a position of no difference or no effect, which we seek to confirm or refute through hypothesis testing.
In the context of an F-test for variances:
  • The null hypothesis states that the variances of the two groups are equal: \( \sigma_1^2 = \sigma_2^2 \).
  • Rejecting the null hypothesis implies there's evidence of a difference in variances.
  • It's the status quo assumption unless evidence strongly suggests otherwise.
Understanding the null hypothesis is crucial as it guides us whether to accept it after comparing the test statistic with the critical values. In our example, we rejected \( H_0 \) because the calculated F was greater than the critical value, suggesting a difference in temperatures' variances.
Alternative Hypothesis
The alternative hypothesis, denoted as \( H_1 \) or sometimes \( H_a \), is the complement of the null hypothesis. It suggests that there is an actual effect or difference, deviating from the status quo.
In our F-test scenario:
  • The alternative hypothesis claims that the two variances are not equal: \( \sigma_1^2 eq \sigma_2^2 \).
  • It represents the research hypothesis or the original claim you set out to prove.
  • The acceptance of the alternative hypothesis relies on the rejection of the null hypothesis.
For this hypothesis, evidence comes from observing an F-test statistic that departs significantly from the critical value range. In our example, the rejection of the null hypothesis led us to accept the alternative hypothesis, indicating a difference in variance between January and February’s temperatures.

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 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. National statistics show that \(23 \%\) of men smoke and \(18.5 \%\) of women smoke. A random sample of 180 men indicated that 50 were smokers, and a random sample of 150 women surveyed indicated that 39 smoked. Construct a \(98 \%\) confidence interval for the true difference in proportions of male and female smokers. Comment on your interval-does it support the claim that there is a difference?

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. A real estate agent compares the selling prices of randomly selected homes in two municipalities in southwestern Pennsylvania to see if there is a difference. The results of the study are shown. Is there enough evidence to reject the claim that the average cost of a home in both locations is the same? Use \(\alpha=0.01\). $$ \begin{array}{ll} {\text { Scott }} & {\text { Ligonier }} \\ \hline \bar{X}_{1}=\$ 93,430^{*} & \bar{X}_{2}=\$ 98,043^{*} \\ \sigma_{1}=\$ 5602 & \sigma_{2}=\$ 4731 \\ n_{1}=35 & n_{2}=40 \end{array} $$

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. A random survey of 80 women who were victims of violence found that 24 were attacked by relatives. A random survey of 50 men found that 6 were attacked by relatives. At \(\alpha=0.10,\) can it be shown that the percentage of women who were attacked by relatives is greater than the percentage of men who were attacked by relatives?

Instead of finding the mean of the differences between \(X_{1}\) and \(X_{2}\) by subtracting \(X_{1}-X_{2}\), you can find it by finding the means of \(X_{1}\) and \(X_{2}\) and then subtracting the means. Show that these two procedures will yield the same results.

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. It has been found that \(26 \%\) of men 20 years and older suffer from hypertension (high blood pressure) and \(31.5 \%\) of women are hypertensive. A random sample of 150 of each gender was selected from recent hospital records, and the following results were obtained. Can you conclude that a higher percentage of women have high blood pressure? Use \(\alpha=0.05\) Men \(\quad 43\) patients had high blood pressure Women 52 patients had high blood pressure

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.