/*! 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 9 Effects of Parental Education on... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 9

Effects of Parental Education on Boys' Education (Example 2) Refer to Exercise 10.7. The data table compares 15 -year-old boys who are either attending school or have dropped out, in order to understand the impact of parental education on them. Report the observed value of the chi-square statistic. \begin{tabular}{|l|c|c|} \hline & Educated Parents & Uneducated Parents \\ \hline Studying & 42 & 23 \\ \hline Not Studying & 14 & 8 \\ \hline \end{tabular}

Short Answer

Expert verified
The observed value of the chi-square statistic for this data is approximately 0.011.

Step by step solution

01

Compute the Marginal Totals

First, compute the totals for each row and each column. Here's the table:\[\begin{tabular}{|l|c|c|c|}\hline & Educated Parents & Uneducated Parents & Total \\hline Studying & 42 & 23 & 65 \\hline Not Studying & 14 & 8 & 22 \\hline Total & 56 & 31 & 87 \\hline\end{tabular}\]
02

Calculate the Expected Frequencies

Next, calculate the expected frequencies for each cell by multiplying the total of the corresponding row by the total of the corresponding column, and then dividing by the grand total. Expected frequency for studying boys with educated parents = \((65 * 56) / 87 = 41.72\)Expected frequency for studying boys with uneducated parents = \((65 * 31) / 87 = 23.28\)Expected frequency for not studying boys with educated parents = \((22 * 56) / 87 = 14.28\)Expected frequency for not studying boys with uneducated parents = \((22 * 31) / 87 = 7.72\)
03

Compute the Chi-Square Statistic

Lastly, use the observed and expected values to compute the chi-square statistic. This is done by summing the squared difference between the observed and expected values, divided by the expected value for each cell. Chi-Square Statistic = \(\sum ((O_i - E_i)^2 / E_i)\)Substituting the observed (O) and expected (E) values,\[\chi^2 = ((42 - 41.72)^2 / 41.72) + ((23 - 23.28)^2 / 23.28) + ((14 - 14.28)^2 / 14.28) + ((8 - 7.72)^2 / 7.72) = 0.011\]
04

Report the Observed Chi-Square Statistic Value

The observed value of the chi-square statistic is 0.011

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.

Parental Education
Parental education plays a significant role in shaping a child's educational journey. In the context of this exercise, we see how it may influence the tendency of boys either to continue their schooling or to drop out at the age of 15. When we refer to 'educated' parents, we generally mean those who have attained a certain level of formal education. Similarly, 'uneducated' parents are those who may not have had access to or completed much formal education.

Understanding these distinctions is crucial because education can impact factors such as family expectations, access to resources, and even the value placed on schooling. Children with educated parents might have more support or encouragement to continue their education, but it's also important to consider that each family and individual is unique. Not all boys with uneducated parents will drop out, as many other variables, such as personal ambition or external support, can come into play.
  • Parental expectations and encouragement
  • Access to educational resources
  • Community and external support systems
Statistical Analysis
Statistical analysis provides the tools necessary to interpret data and glean insights, like the effects of parental education on schooling. In this exercise, statistical methods help us assess patterns and relationships within the data.

Using a Chi-Square test, we aim to determine if there's a statistically significant association between parental education levels and the educational status of boys. The Chi-Square test is particularly suited for categorical data like this, where we examine frequencies across different groups.
  • Helps to decide if observed differences are meaningful
  • Analyses categorical data effectively
  • Expresses statistical significance through a p-value

The process involves calculating how likely the observed data would happen under the assumption that there's no association (null hypothesis). A small Chi-Square statistic indicates no strong relationship, while a large statistic suggests a potential link. In this case, a Chi-Square value of 0.011 suggests that there's little to no significant association between parental education and whether boys continue school.
Expected Frequencies
Expected frequencies provide a benchmark to compare what we observe against what we expect if there's no association between variables. To calculate these, the marginal totals of rows and columns are crucial. They tell us how many occurrences we would anticipate in each category if parental education didn't influence schooling outcomes at all.

The process involves multiplying the corresponding row total by the column total, and then dividing by the overall total. For instance, expected frequencies help us understand the distribution we would expect,
  • Compute using row and column totals
  • Provides a model for comparison
  • Helps in chi-square calculation
If the observed frequencies significantly deviate from these expected numbers, it suggests that factors other than mere chance are at play. However, in this example, calculated expected frequencies roughly match the observed data, reinforcing a minimal association, as indicated by the Chi-Square statistic.

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

Removal of Healthy Appendixes Computed tomography (CT) scans are used to diagnose the need for the removal of the appendix. CT scans give the patient a large level of radiation, which has risks, especially for young people. There is a new form of \(\mathrm{CT}\) scanning called low-dose CT, which was tested to see whether it was inferior when diagnosing appendicitis. Negative appendectomies are appendectomies that were done even though the appendix was healthy. The negative appendectomy rate was 6 of 172 patients randomly assigned to the low-dose \(C T\) and 6 out of 186 patients randomly assigned to the standard-dose group. a. Find the negative appendectomy rates for both samples and compare them. b. Test the hypothesis that the negative appendectomy rate and dosage are independent at the \(0.05\) level.

Steroids and Height Does the use of inhaled steroids by children affect their height as adults? Excerpts from the abstract of a study about this are given. Read them and then answer the questions that follow. "Methods: We measured adult height in 943 of 1041 participants \((90.6 \%)\) in the Childhood Asthma Management Program; adult height was determined at a mean ( \(\pm \mathrm{SD}\) ) age of \(24.9 \pm 2.7\) years. Starting at the age of 5 to 13 years, the participants had been randomly assigned to receive \(400 \mu \mathrm{g}\) of budesonide, 16 \(\mathrm{mg}\) of nedocromil, or placebo daily for 4 to 6 years. Results: Mean adult height was \(1.2 \overline{\mathrm{cm}}\) lower \((95 \%\) confidence interval \([\mathrm{CI}],-1.9\) to \(-0.5\) ) in the budesonide group than in the placebo group \((\mathrm{P}=0.001)\) and was \(0.2 \mathrm{~cm}\) lower (95\% CI, \(-0.9\) to \(0.5\) ) in the nedocromil group than in the placebo group \((\mathrm{P}=0.61) . "\) a. Identify the treatment variable and the response variable. b. Was this a controlled experiment or an observational study? Explain. c. Does the first interval, \((-1.9\) to \(-0.5\) ), capture 0 ? What does that show? d. From the interval, can you conclude that the use of budesonide in childhood reduces the heights of the children when they become adults? Why or why not? e. Does the second interval, \((-0.9\) to \(0.5\) ), capture 0 ? What does that show? f. From the interval, can you conclude that the use of nedocromil in childhood reduces the heights of the children when they become adults? Why or why not?

In Montreal, Canada, an experiment was done with parents of children who were thought to have a high risk of committing crimes when they became teenagers. Some of the families were randomly assigned to receive parental training, and the others were not. Out of 43 children whose parents were randomly assigned to the parental training group, 6 had been arrested by the age of \(15 .\) Out of 123 children whose parents were not in the parental training group, 37 had been arrested by age 15 . a. Find and compare the percentages of children arrested by age \(15 .\) Is this what researchers might have hoped? b. Create a two-way table from the data, and test whether the treatment program is associated with arrests. Use a significance level of \(0.05\). c. Do a two-proportion \(z\) -test, testing whether the parental training lowers the rate of bad results. Use a significance level of \(0.05\). d. Explain the difference in the results of the chi-square test and the two- proportion z-test. e. Can you conclude that the treatment causes the better result? Why or why not?

Odd-Even Formula A survey was taken of a random sample of people noting their gender and asking whether they agreed with the Odd-Even Formula (OEF) to control the alarming levels of air pollution. Minitab results are shown. \(=1\) 2 Chi-square Test for Association: Opinion, Gender ciat:1 Rows: Ooinion Columns: Gender $$ \begin{array}{rr}\text { Male Female } \\ \text { Disagree } & 42 & 44 \\\ 42.17 & 43.83 \\ \text { Agree } & 11.4 & 86 \\ & 110 & 14.17 & \\ & 109.83 & \\ \mathrm{All} & & \\ & 152 & 158 & 310 \\ \text { Cel1 Contents: } & \text { Count } \\ & \text { Expected count } \\ \text { Pearson Chi-Square }=0.002, \mathrm{DF}=1, \text { p-value }=0.966\end{array} $$ a. Find the percentage of men and women in the sample who agreed with the OEF method, and compare these percentages. b. Test the hypothesis that opinions about OEF and gender are independent using a significance level of \(0.05\). c. Does this suggest that men and women have significantly different views about the OEF method?

Antiretrovirals to Prevent HIV A study conducted in Uganda and Kenya looked at heterosexual couples in which one of the partners was HIV-positive and the other was not. The person in each couple who was not HIV-positive was randomly assigned to one of three study regimens: tenofovir (TDF), combination tenofovir-emtricitabine (TDF-FTC), or placebo and was followed for up to 36 months. Seventeen of the 1584 people assigned to TDF became positive for HIV, as did 13 of the 1579 assigned to TDFFTC and 52 of the 1584 assigned to the placebo. a. Find the percentage in cach group in the sample that became HIV positive, and compare these percentages. b. Create a two-way table with the treatment labels across the top. c. Test the hypothesis that treatment and HIV status are associated using a significance level of \(0.05\). (Source: J. Baelen et al. 2012. Antiretroviral prophylaxis for HIV prevention in heterosexual men and women. New England Journal of Medicine \(367,399-410\), August.)
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Logic and Functions

Read Explanation

Discrete Mathematics

Read Explanation

Geometry

Read Explanation

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

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.