/*! 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 10 Determine \((a)\) the \(\chi^{2}... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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 \(n=4, p=0.3\) \(H_{1}:\) The random variable \(X\) is not binomial with \(n=4\) \(p=0.3\) $$ \begin{array}{llllll} \boldsymbol{X} & \mathbf{0} & \mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4} \\ \hline \text { Observed } & 260 & 400 & 280 & 50 & 10 \\ \hline \text { Expected } & 240.1 & 411.6 & 264.6 & 75.6 & 8.1 \\ \hline \end{array} $$

Short Answer

Expert verified
The chi-square test statistic is 11.645. Degrees of freedom are 4. Critical value at \( \alpha = 0.05 \) is 9.488. Reject \( H_0 \).

Step by step solution

01

- Calculate the Chi-Square Test Statistic

The Chi-Square test statistic \(n \chi^{2} \) is calculated using the formula \[ \chi^{2} = \sum \frac{(O_i - E_i)^2}{E_i} \] where \( O_i \) are the observed frequencies and \( E_i \) are the expected frequencies. Compute this for each value in the table: \(\chi^{2} = \frac{(260 - 240.1)^2}{240.1} + \frac{(400 - 411.6)^2}{411.6} + \frac{(280 - 264.6)^2}{264.6} + \frac{(50 - 75.6)^2}{75.6} + \frac{(10 - 8.1)^2}{8.1}\) Calculate each term separately: \[ \chi^{2}_0 = \frac{(260 - 240.1)^2}{240.1} \approx 1.656 \] \[ \chi^{2}_1 = \frac{(400 - 411.6)^2}{411.6} \approx 0.328 \] \[ \chi^{2}_2 = \frac{(280 - 264.6)^2}{264.6} \approx 0.854 \] \[ \chi^{2}_3 = \frac{(50 - 75.6)^2}{75.6} \approx 8.379 \] \[ \chi^{2}_4 = \frac{(10 - 8.1)^2}{8.1} \approx 0.428 \] Sum these up: \[ \chi^{2} \approx 11.645 \]
02

- Determine the Degrees of Freedom

Degrees of freedom \( df \) are calculated as the number of categories minus 1. Here, there are 5 categories (0, 1, 2, 3, 4): \[ df = 5 - 1 = 4 \]
03

- Find the Critical Value

Using a chi-square distribution table, find the critical value for \( df = 4 \) at \(n \alpha = 0.05 \). The critical value is \[ \chi^{2}_{critical} = 9.488 \]
04

- Test the Hypothesis

Compare the calculated chi-square test statistic (Step 1) with the critical value (Step 3). \[ \chi^{2}_{calculated} = 11.645 \]\[ \chi^{2}_{critical} = 9.488 \] Since \( 11.645 > 9.488 \), reject the null hypothesis \( H_0 \). There is sufficient evidence to conclude that the random variable \( X \) is not binomial with \( n=4 \) and \( p=0.3 \).

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.

Degrees of Freedom
Degrees of freedom (df) are a key concept in various statistical tests, including the Chi-Square test. Degrees of freedom generally refer to the number of independent values or quantities which can be assigned to a statistical distribution. In the context of the Chi-Square test, degrees of freedom are calculated as the number of categories minus 1. This accounts for the fact that if you know the total number of observations, and you know the counts in all but one category, then the count in the remaining category is fixed. For example, in our given table with 5 categories (0, 1, 2, 3, 4), the degrees of freedom are:
  • df = 5 - 1
Therefore, we get:
  • df = 4
These 4 degrees of freedom mean that 4 categories can vary independently, providing the flexibility needed for the chi-square test calculations. Understanding degrees of freedom helps in grasping the underlying mechanics of the Chi-Square test and other related statistical methods.
Critical Value
The critical value in hypothesis testing is a threshold that distinguishes whether your test statistic falls in the rejection region (leading to rejecting the null hypothesis) or not. For the Chi-Square test, the critical value is found using a Chi-Square distribution table at a predetermined significance level ( (alpha) ). In this exercise, our significance level is (alpha = 0.05). To find the critical value:
  • Determine the degrees of freedom.
  • Locate the critical value in a Chi-Square distribution table corresponding to (degr=f = 4) and (degree = 0.05).
In our case, the critical value at (df = 4) and (alpha = 0.05) is:
  • ( ( Chi^2_{critical} = 9.488)
This means that if our calculated Chi-Square test statistic is greater than 9.488, we reject the null hypothesis.
Hypothesis Testing
Hypothesis testing is a method used to determine if there is enough statistical evidence in favor of a specific hypothesis. In this exercise, we are testing the claim (null hypothesis (H_0)) that the random variable (X) follows a binomial distribution with parameters ( = 4) and (p=0.3). The steps for hypothesis testing include:
  • State the null and alternate hypotheses: (H_0) and (H_1)).
  • Choose a significance level, (alpha = 0.05) in this case.
  • Compute the test statistic (Using Chi-Square in this case).
  • Determine the critical value.
  • Compare the calculated test statistic to the critical value.
  • Make a decision: reject or fail to reject the null hypothesis ((H_0)).
For our exercise, we calculated the Chi-Square test statistic to be approximately 11.645. Our critical value was found to be 9.488. Since 11.645 is greater than 9.488, we reject (H_0), supporting the alternate hypothesis (H_1) that (X) is not binomial with (n = 4) and (p = 0.3).
Statistical Significance
Statistical significance is a determination of whether the result of a hypothesis test is extreme enough to reject the null hypothesis. It relates to the (alpha) level, which is the threshold for deciding significance. In this exercise, the (degree = 0.05) level indicates that there's a 5% risk of concluding that a difference exists when there is no actual difference.
  • If the test statistic exceeds the critical value derived from the Chi-Square distribution for given degrees of freedom, it indicates statistical significance.
Our test statistic was 11.645, which exceeded our critical value of 9.488 at <(alpha = 0.05).) This means our result is statistically significant. Hence, we reject the null hypothesis (H_0) and accept that the data does not fit a binomial distribution with the provided parameters. Understanding statistical significance helps in making informed decisions in hypothesis testing and interpreting test results.

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

Family Structure and Sexual Activity A sociologist wants to discover whether the sexual activity of females between the ages of 15 and 19 years and family structure are associated. She randomly selects 380 females between the ages of 15 and 19 years and asks each to disclose her family structure at age 14 and whether she has had sexual intercourse. The results are shown in the table. Data are based on information obtained from the National Center for Health Statistics. $$ \begin{array}{lcccc} &&{\text { Family Structure }} \\ \hline & \text { Both Biological } & & & \\ \text { Had Sexual } & \text { or Adoptive } & \text { Single } & \text { Parent and } & \text { Nonparental } \\ \text { Intercourse } & \text { Parents } & \text { Parent } & \text { Stepparent } & \text { Guardian } \\ \hline \text { Yes } & 64 & 59 & 44 & 32 \\ \hline \text { No } & 86 & 41 & 36 & 18 \\ \hline \end{array} $$ (a) Compute the expected values of each cell under the assumption of independence. (b) Verify that the requirements for performing a chi-square test of independence are satisfied. (c) Compute the chi-square test statistic. (d) Test whether family structure and sexual activity of 15 - to 19-year-old females are independent at the \(\alpha=0.05\) level of significance. (e) Compare the observed frequencies with the expected frequencies. Which cell contributed most to the test statistic? Was the expected frequency greater than or less than the observed frequency? What does this information tell you? (f) Construct a conditional distribution by family structure and draw a bar graph. Does this evidence support your conclusion in part (d)?

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. $$ \begin{array}{lcccc} \hline\text { Outcome } & \mathbf{A} & \mathbf{B} & \mathbf{C} & \mathbf{D} \\\ \hline \text { Observed } & 30 & 20 & 28 & 22 \\ \hline \text { Expected } & 25 & 25 & 25 & 25 \\ \hline \end{array} $$

Does the location of your seat in a classroom play a role in attendance or grade? To answer this question, professors randomly assigned 400 students a general education physics course to one of four groups. The 100 students in group 1 sat 0 to 4 meters from the front of the class, the 100 students in group 2 sat 4 to 6.5 meters from the front, the 100 students in group 3 sat 6.5 to 9 meters from the front, and the 100 students in group 4 sat 9 to 12 meters from the front. (a) For the first half of the semester, the attendance for the whole class averaged \(83 \% .\) So, if there is no effect due to seat location, we would expect \(83 \%\) of students in each group to attend. The data show the attendance history for each group. How many students in each group attended, on average? Is there a significant difference among the groups in attendance patterns? Use the \(\alpha=0.05\) level of significance. (b) For the second half of the semester, the groups were rotated so that group 1 students moved to the back of class and group 4 students moved to the front. The same switch took place between groups 2 and \(3 .\) The attendance for the second half of the semester averaged \(80 \% .\) The data show the attendance records for the original groups (group 1 is now in back, group 2 is 6.5 to 9 meters from the front, and so on ). How many students in each group attended, on average? Is there a significant difference in attendance patterns? Use the \(\alpha=0.05\) level of significance. Do you find anything curious about these data? $$ \begin{array}{lllll} \hline \text { Group } & 1 & 2 & 3 & 4 \\ \hline \text { Attendance } & 0.84 & 0.81 & 0.78 & 0.76\\\ \hline \end{array} $$ (c) At the end of the semester, the proportion of students in the top \(20 \%\) of the class was determined. Of the students in group \(1,25 \%\) were in the top \(20 \%\); of the students in group \(2,21 \%\) were in the top \(20 \%\); of the students in group \(3,15 \%\) were in the top \(20 \%\); of the students in group \(4,19 \%\) were in the top \(20 \% .\) How many students would we expect to be in the top \(20 \%\) of the class if seat location plays no role in grades? Is there a significant difference in the number of students in the top \(20 \%\) of the class by group? (d) In earlier sections, we discussed results that were statistically significant, but did not have any practical significance. Discuss the practical significance of these results. In other words, given the choice, would you prefer sitting in the front or back?

True or False: The expected frequencies in a chi-square test for independence are found using the formula Expected frequency \(=\frac{(\text { row total })(\text { column total })}{\text { table total }}\)

Determine the expected counts for each outcome. $$ \begin{array}{lllll} \hline \boldsymbol{n}=\mathbf{5 0 0} & & & & \\ \hline p_{i} & 0.2 & 0.1 & 0.45 & 0.25 \\ \hline \text { Expected counts } & & & & \\ \hline \end{array} $$

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.