/*! 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 7 Determine \((a)\) the \(\chi^{2}... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 7

Determine \((a)\) the \(\chi^{2}\) test statistic, \((b)\) the degrees of freedom, (c) the critical value using \(\alpha=0.05,\) and (d) test the hypothesis at the \(\alpha=0.05\) level of significance. \(H_{0}: p_{\mathrm{A}}=p_{\mathrm{B}}=p_{\mathrm{C}}=p_{\mathrm{D}}=\frac{1}{4}\) \(H_{1}\) : At least one of the proportions is different from the others. $$ \begin{array}{lcccc} \hline\text { Outcome } & \mathbf{A} & \mathbf{B} & \mathbf{C} & \mathbf{D} \\\ \hline \text { Observed } & 30 & 20 & 28 & 22 \\ \hline \text { Expected } & 25 & 25 & 25 & 25 \\ \hline \end{array} $$

Short Answer

Expert verified
Test statistic: 2.72, degrees of freedom: 3, critical value: 7.815. We fail to reject the null hypothesis.

Step by step solution

01

Calculate the Chi-Square Test Statistic

Use the formula for the chi-square test statistic: \[\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}\]Where \(O_i\) is the observed frequency and \(E_i\) is the expected frequency. For each category A, B, C, and D, calculate the terms:\[\frac{(30 - 25)^2}{25} = 1,\frac{(20 - 25)^2}{25} = 1,\frac{(28 - 25)^2}{25} = 0.36,\frac{(22 - 25)^2}{25} = 0.36\]Sum these up to get the \(\chi^2\) test statistic:\[\chi^2 = 1 + 1 + 0.36 + 0.36 = 2.72\]
02

Determine Degrees of Freedom

The degrees of freedom (df) for a chi-square test are calculated with the formula: \[df = (k - 1)\]Where \(k\) is the number of categories. Here, there are 4 categories (A, B, C, D), so the degrees of freedom are:\[df = 4 - 1 = 3\]
03

Find the Critical Value

Using the chi-square distribution table, find the critical value for \(\alpha = 0.05\) and \(df = 3\). The critical value is 7.815.
04

Compare Test Statistic with Critical Value

Compare the calculated \(\chi^2\) test statistic (2.72) with the critical value (7.815). Since 2.72 < 7.815, we fail to reject the null hypothesis.
05

Conclusion

At the \(\alpha = 0.05\) level of significance, the test does not provide sufficient evidence to reject the null hypothesis. Therefore, we conclude that there is no significant difference between the observed and expected frequencies.

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.

degrees of freedom
Degrees of freedom (df) are an essential concept in statistics, particularly in hypothesis testing. They refer to the number of independent values or quantities that can vary in a statistical calculation. In the context of a chi-square test, degrees of freedom help determine the shape of the chi-square distribution, which in turn helps find the critical value needed for decision-making.

The formula for calculating degrees of freedom in a chi-square test is straightforward:\[df = (k - 1)\]where \(k\) is the number of categories being compared. For example, if you have 4 categories (A, B, C, D), the degrees of freedom would be: \[df = 4 - 1 = 3\].
Degrees of freedom are crucial because they influence the critical value that you will look up in chi-square distribution tables. The higher the degrees of freedom, the more the chi-square distribution shifts and expands.
critical value
The critical value is a threshold used in statistical tests to decide whether to accept or reject the null hypothesis. It is based on the desired significance level (\(\alpha\)), which is usually set at 0.05 for a 95% confidence level. For a chi-square test, the critical value is found in the chi-square distribution table and is determined by the degrees of freedom and \(\alpha\).

For instance, let's say you have 3 degrees of freedom and an \(\alpha\) of 0.05. To find the critical value, you would look at the chi-square distribution table under these parameters. In our example, the critical value is 7.815.

Comparing the calculated chi-square test statistic (2.72) to this critical value (7.815) can help you make a conclusion. If your test statistic is less than the critical value, you fail to reject the null hypothesis; if it is greater, you reject the null hypothesis.
null hypothesis
The null hypothesis (\(H_0\)) is a fundamental aspect of statistical testing. It represents a statement of no effect or no difference. For a chi-square test, the null hypothesis often states that the observed frequencies are equal to the expected frequencies, meaning that any differences are due to random chance.

In our example, the null hypothesis is: \[ H_0: p_A = p_B = p_C = p_D = \frac{1}{4} \].
This means we assume that the proportions across categories A, B, C, and D are equal unless proven otherwise.
Alternatively, the alternative hypothesis (\(H_1\)) states that at least one of these proportions is different. After calculating the chi-square test statistic and the critical value, comparing them helps decide whether to reject the null hypothesis. If the test statistic is less than the critical value, it means we don't have enough evidence to reject \(H_0\), so we assume that the observed frequencies match the expected frequencies.

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

On February 2,1894 Frank Raphael Weldon wrote a letter to Francis Galton that included the results of 26,306 rolls of 12 dice. Weldon recorded the results such that a roll of a 5 or 6 resulted in a success, while a roll of \(1,2,3,\) or 4 was a failure. The number of successes in each roll of the 12 dice are recorded in the table. $$ \begin{array}{cc|cc} \begin{array}{l} \text { Number of } \\ \text { Successes } \end{array} & \text { Frequency } & \begin{array}{l} \text { Number of } \\ \text { Successes } \end{array} & \text { Frequency } \\ \hline 0 & 185 & 7 & 1331 \\ \hline 1 & 1149 & 8 & 403 \\ \hline 2 & 3265 & 9 & 105 \\ \hline 3 & 5475 & 10 & 14 \\ \hline 4 & 6114 & 11 & 4 \\ \hline 5 & 5194 & 12 & 0 \\ \hline 6 & 3067 & & \\ \hline \end{array} $$ (a) What is the probability of rolling a 5 or 6 when throwing a six-sided fair die? (b) Treating the probability determined in part (a) as the probability of success, compute the theoretical probability of \(0,1,2, \ldots, 12\) successes in throwing 12 dice. (c) Use the probabilities found in part (b) to determine the expected frequency in observing \(0,1,2, \ldots, 12\) successes after throwing the 12 dice 26,306 times. (d) Conduct a goodness-of-fit test to determine if the number of successes follows a binomial probability distribution. Note: Combine 11 and 12 into a single bin.

In a survey of 3029 adult Americans, the Harris Poll asked people whether they smoked cigarettes and whether they always wear a seat belt in a car. The table shows the results of the survey. For each activity, we define a success as finding an individual who participates in the hazardous activity. $$ \begin{array}{lcc} & \begin{array}{c} \text { No Seat Belt } \\ \text { (success) } \end{array} & \begin{array}{c} \text { Seat Belt } \\ \text { (failure) } \end{array} \\ \hline \text { Smoke (success) } & 67 & 448 \\ \hline \text { Do not smoke (failure) } & 327 & 2187 \\ \hline \end{array} $$ (a) Why is this a dependent sample? (b) Is there a significant difference in the proportion of individuals who smoke and the proportion of individuals who do not wear a seat belt? In other words, is there a significant difference between the proportion of individuals who engage in hazardous activities? Use the \(\alpha=0.05\) level of significance.

Determine \((a)\) the \(\chi^{2}\) test statistic, \((b)\) the degrees of freedom, (c) the critical value using \(\alpha=0.05,\) and (d) test the hypothesis at the \(\alpha=0.05\) level of significance. \(H_{0}:\) The random variable \(X\) is binomial with \(n=4, p=0.8\) \(H_{1}:\) The random variable \(X\) is not binomial with \(n=4\) \(p=0.8\) $$ \begin{array}{llllll} \boldsymbol{X} & \mathbf{0} & \mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4} \\ \hline \text { Observed } & 1 & 38 & 132 & 440 & 389 \\ \hline \text { Expected } & 1.6 & 25.6 & 153.6 & 409.6 & 409.6 \\ \hline \end{array} $$

The following table contains observed values and expected values in parentheses for two categorical variables, \(X\) and \(Y\), where variable \(X\) has three categories and variable \(Y\) has two categories: $$ \begin{array}{cccc} & \boldsymbol{X}_{\mathbf{1}} & \boldsymbol{X}_{\mathbf{2}} & \boldsymbol{X}_{3} \\ \hline \boldsymbol{Y}_{\mathbf{1}} & 34(36.26) & 43(44.63) & 52(48.11) \\ \hline \boldsymbol{Y}_{\mathbf{2}} & 18(15.74) & 21(19.37) & 17(20.89) \\ \hline \end{array} $$ (a) Compute the value of the chi-square test statistic. (b) Test the hypothesis that \(X\) and \(Y\) are independent at the \(\alpha=0.05\) level of significance.

How much does the typical person pay for a new 2015 Buick Regal? The following data represent the selling price of a random sample of new Regals (in dollars). $$ \begin{array}{lllll} \hline 41,215 & 41,303 & 41,453 & 41,898 & 40,988 \\ \hline 40,078 & 41,215 & 39,623 & 42,352 & 41,898 \\ \hline 40,533 & 42,580 & 40,306 & 41,670 & 39,851 \end{array} $$ (a) Is this data quantitative or qualitative? (b) Find the mean and median price of a new 2015 Regal. (c) Find the standard deviation and interquartile range. (d) Verify it is reasonable to conclude that this data come from a population that is normally distributed. (e) Draw a boxplot of the data. (f) Estimate the typical price paid for a new 2015 Buick Regal with \(90 \%\) confidence. (g) Would a \(90 \%\) confidence interval for all new 2015 domestic vehicles be wider or narrower? Explain.
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Statistics

Read Explanation

Decision Maths

Read Explanation

Logic and Functions

Read Explanation

Mechanics Maths

Read Explanation

Discrete Mathematics

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.