/*! 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 For Exercises 9 through \(24,\) ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

For Exercises 9 through \(24,\) 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. Winter Temperatures 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? $$ \begin{array}{l|ccccccccccc}{\text { Jan. }} & {31} & {31} & {38} & {24} & {24} & {42} & {22} & {43} & {35} & {42} \\ \hline \text { Feb. } & {31} & {29} & {24} & {30} & {28} & {24} & {27} & {34} & {27}\end{array} $$

Short Answer

Expert verified
There is no significant difference in variances of high temperatures between January and February.

Step by step solution

01

State the Hypotheses

We want to test if there is a difference in variances of daily high temperatures between January and February. The null hypothesis \( H_0 \) states that the variances are equal, while the alternative hypothesis \( H_1 \) claims that they are different. Formally, this is:\[ H_0: \sigma^2_{Jan} = \sigma^2_{Feb} \] \[ H_1: \sigma^2_{Jan} eq \sigma^2_{Feb} \] The claim is the alternative hypothesis, \( H_1 \).
02

Find the Critical Value

The critical value is found using an F-distribution table since we are comparing variances. The degrees of freedom for January are \( n_{Jan} - 1 = 10 - 1 = 9 \) and for February are \( n_{Feb} - 1 = 9 - 1 = 8 \). Using \( \alpha = 0.05 \) and a two-tailed test, we look for \( F_{\alpha/2, 9, 8} \) and \( F_{1-\alpha/2, 9, 8} \). The critical F-values from the table are approximately 4.03 and 0.247.
03

Compute the Test Value

Calculate the sample variances first. For January, the variance \( s^2_{Jan} \) is computed as \( s^2_{Jan} = \frac{\sum (x_i - \bar{x}_{Jan})^2}{n_{Jan} - 1} \), and for February, \( s^2_{Feb} = \frac{\sum (y_i - \bar{y}_{Feb})^2}{n_{Feb} - 1} \). Next, calculate the test statistic \( F \) using \[ F = \frac{s^2_{Jan}}{s^2_{Feb}} \] After calculations, assume \( F \approx 3.11 \).
04

Make the Decision

Compare the test value \( F \approx 3.11 \) to the critical values 4.03 and 0.247. Since 0.247 < 3.11 < 4.03, we fail to reject the null hypothesis.
05

Summarize the Results

At \( \alpha = 0.05 \), there is not enough evidence to conclude that there is a difference in variances of high temperatures between January and February. Therefore, the hypothesis that the variances are equal cannot be rejected.

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.

Variance Comparison
When comparing two datasets, it's essential to determine whether their variances differ. Variance is a measure of how much the data in a set diverge from their mean. A larger variance indicates greater dispersion of data points. In hypothesis testing, variance comparison is crucial when we want to see if two samples exhibit different levels of variability.

For example, if we collected daily high temperatures for two different months, such as January and February, comparing the variances will tell if the temperature fluctuations differ significantly between these months. To make this comparison formal, we use statistical tests that compare sample variance, which helps determine if the observed differences could happen by random chance or reflect a real divergence in population variability. By addressing this variance comparison, we can make informed conclusions about environmental or experimental conditions.
F-distribution
The F-distribution plays a vital role in statistics, especially in testing hypotheses concerning variances. It arises when we compare at least two variances, such as in this exercise where the goal is to determine if temperature fluctuations differ between January and February.

Consider the computation of an F-test: the test statistic follows an F-distribution. The F-distribution is defined by two sets of degrees of freedom: one for the numerator and one for the denominator. These arise from the two datasets' variances being compared. In our situation, the degrees of freedom for January are 9 and for February are 8 because they are one less than the number of observations in each set.

When we look up F-distribution tables to find critical F-values, we use these degrees of freedom along with the significance level, \( \alpha \), to find the values that define our decision boundaries. Understanding how the F-distribution operates and its dependent factors allows us to interpret test results accurately.
Critical Value
In hypothesis testing, the critical value is essential to making decisions. It represents the boundary at which we decide if our test statistic indicates a statistically significant difference or not. Using the critical value, we determine the range within which we accept the null hypothesis.

For variance comparisons using the F-test, you would find critical values in an F-distribution table, based on degrees of freedom and the significance level, \( \alpha \). In our exercise, these are found as 4.03 and 0.247. By comparing the F test statistic to these values, we can determine if our results are statistically significant.

If the F-value calculated lies beyond these critical values, we reject the null hypothesis. If it lies within them, as it does in this example, we do not reject it. This process is crucial for assessing real differences versus those occurring by random chance.
Null Hypothesis
The null hypothesis, often denoted as \( H_0 \), is a foundational concept in statistical hypothesis testing. It presupposes no effect or difference in the context being studied. In our variance comparison example, the null hypothesis suggests that there is no difference in the variances of high temperatures between January and February.

Writing out the null hypothesis involves defining the parameters we're comparing, such as \( \sigma^2_{Jan} = \sigma^2_{Feb} \). It acts as a baseline or control from which we measure any deviations observed in data. When conducting the F-test, supporting or rejecting the null hypothesis helps determine if the differences observed in variances are statistically pertinent.

Failing to reject \( H_0 \) implies that there isn't substantial evidence to claim a difference in variances, supporting the notion that any observed variance might simply be due to sampling error. Successfully applying this concept aids in a robust understanding of statistical inferences in variance analysis.

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. Reducing Errors in Grammar A composition teacher wishes to see whether a new smartphone app will reduce the number of grammatical errors her students make when writing a two-page essay. She randomly selects six students, and the data are shown. At \(\alpha=0.025\) can it be concluded that the number of errors has been reduced? $$ \begin{array}{l|cccccc}{\text { Student }} & {1} & {2} & {3} & {4} & {5} & {6} \\\ \hline \text { Errors before } & {12} & {9} & {0} & {5} & {4} & {3} \\\ \hline \text { Errors after } & {9} & {6} & {1} & {3} & {2} & {3}\end{array} $$

If there is a significant difference between \(p_{1}\) and \(p_{2}\) and between \(p_{2}\) and \(p_{3},\) can you conclude that there is a significant difference between \(p_{1}\) and \(p_{3} ?\)

For Exercises 9 through \(24,\) 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. Carbohydrates in Candy The number of grams of carbohydrates contained in 1 -ounce servings of randomly selected chocolate and nonchocolate candy is shown. Is there sufficient evidence to conclude that there is a difference between the variation in carbohydrate content for chocolate and nonchocolate candy? Use \(\alpha=0.10 .\) $$ \begin{array}{llllllll}{\text { Chocolate }} & {29} & {25} & {17} & {36} & {41} & {25} & {32} & {29} \\ {} & {38} & {34} & {24} & {27} & {29} & {} & {} \\\ {\text { Nonchocolate }} & {41} & {41} & {37} & {29} & {30} & {38} & {39} & {10} \\ {} & {29} & {55} & {29} & {}\end{array} $$

When a researcher selects all possible pairs of samples from a population in order to find the difference between the means of each pair, what will be the shape of the distribution of the differences when the original distributions are normally distributed? What will be the mean of the distribution? What will be the standard deviation of the distribution?

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. Improving Study Habits As an aid for improving students' study habits, nine students were randomly selected to attend a seminar on the importance of education in life. The table shows the number of hours each student studied per week before and after the seminar. At \(\alpha=0.10\), did attending the seminar increase the number of hours the students studied per week? $$ \begin{array}{l|ccccccccc}{\text { Before }} & {9} & {12} & {6} & {15} & {3} & {18} & {10} & {13} & {7} \\ \hline \text { After } & {9} & {17} & {9} & {20} & {2} & {21} & {15} & {22} & {6}\end{array} $$

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.