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A vaccine is available to prevent the contraction of human papillomavirus (HPV). The Centers for Disease Control and Prevention recommends this vaccination for all young girls in two doses. In a 2015 study reported in the Journal of American College Health, Lee et al. studied vaccination rates among Asian American and Pacific Islander (AAPI) women and non-Latina white women. Data are shown in the table. Test the hypothesis that vaccination rates and race are associated. Use a \(0.05\) significance level. $$ \begin{array}{|lcc|} \hline \text { Completed HPV vaccinations } & \text { AAPI } & \text { White } \\\ \hline \text { Yes } & 136 & 1170 \\ \hline \text { No } & 216 & 759 \\ \hline \end{array} $$

Short Answer

Expert verified
Yes, there is statistically significant evidence at the \(0.05\) level to reject the null hypothesis and conclude that vaccination rates and race are associated.

Step by step solution

01

Formulate the hypotheses

The null hypothesis (\(H_0\)) would be 'Vaccination rates and race are not associated', while the alternative hypothesis (\(H_a\)) would be 'Vaccination rates and race are associated'. You judge this based on a significance level of \(0.05\).
02

Calculate the Observed Chi-Square

Chi-Square formula: \[X^2 = \sum \frac{(O-E)^2}{E}\] where O is the observed frequency and E is the expected frequency. For this exercise, calculation for Expected values using this formula: \(E = \frac{(Row Total * Column Total)}{Sample Size}\) Let's calculate the expected values for all cells: For 'AAPI, Yes' cell: \(E = \frac{(136+1170) * (136+216)}{136+1170+216+759} = 304.89\) For 'AAPI, No' cell: \(E = \frac{(136+1170) * (216+759)}{136+1170+216+759} = 47.11\) For 'White, Yes' cell: \(E = \frac{(216+759) * (136+1170)}{136+1170+216+759} = 1001.11\) For 'White, No' cell: \(E = \frac{(216+759) * (216+759)}{136+1170+216+759} = 973.89\) By placing them in the chi-square formula: \(X^2 = \frac{(136-304.89)^2}{304.89} + \frac{(216-47.11)^2}{47.11} + \frac{(1170-1001.11)^2}{1001.11} + \frac{(759-973.89)^2}{973.89} = 87.50\)
03

Find the Critical Chi-Square for \(0.05\) significance level

For a \(0.05\) significance level and degree of freedom equal to \(1\), the critical Chi-square is \(3.841\).
04

Compare the Observed and Critical Chi-Square

If the Observed Chi-Square is greater than the Critical Chi-Square, the null hypothesis is rejected. Here, the Observed Chi-Square (\(87.50\)) is much larger than the Critical Chi-Square (\(3.841\)), so the null hypothesis is rejected, and the statement that vaccination rates and race are associated is deemed to be statistically significant.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hypothesis Testing
In statistics, hypothesis testing is a method used to make decisions or inferences about a population based on a sample of data. The main goal here is to determine whether there is enough evidence to reject a null hypothesis, which assumes there is no effect or association between the variables in question.
When performing a hypothesis test, you generally start with two hypotheses:
  • The null hypothesis ( H_0 ), which in this case, states that there is no association between vaccination rates and race. This is what you assume to be true until proven otherwise.
  • The alternative hypothesis ( H_a ), which suggests that there is an association. It's what you want to prove is true.
The hypothesis test involves collecting data and determining the probability of observing that data if the null hypothesis were true. If this probability is sufficiently low, you reject the null hypothesis. This process helps researchers understand if their findings reflect true effects rather than random variation.
Significance Level
The significance level, often denoted as \\( \alpha \)\, is a critical component of hypothesis testing. It represents the probability of rejecting the null hypothesis when it is actually true, also known as making a Type I error. Common significance levels are 0.05, 0.01, and 0.10.
In this exercise, a significance level of 0.05 was used. This means there's a 5% risk that the results are due to random chance rather than a true effect. A lower significance level would mean you're being more conservative, requiring stronger evidence to reject the null hypothesis.This threshold acts as a decision rule and helps statisticians and researchers maintain control over the likelihood of drawing incorrect conclusions from their studies. Moreover, it provides a clear guideline for interpreting results, particularly when comparing observed statistical measures to critical values from appropriate distributions.
Expected Frequency
Expected frequencies play a vital role in the Chi-Square Test. They are calculations of how the data might look if the null hypothesis were true. In other words, they are the frequencies you would expect to find in each cell of a contingency table if there were no association between the variables.
The formula for calculating expected frequency is: \\( E = \frac{(\text{Row Total} \times \text{Column Total})}{\text{Sample Size}} \).
In our solution, the expected frequencies for different cells such as 'AAPI, Yes', and 'White, No' were calculated. This involves using the row and column totals of the table, along with the overall sample size, which gives you a theoretical model to compare against the actual observed data.These expected frequencies are crucial for computing the Chi-Square statistic, as they serve as a benchmark to assess the differences between the observed and expected outcomes. The closer the observed frequencies are to the expected frequencies, the less likely it is that the variables are associated.
Observed Frequency
Observed frequencies refer to the actual counts you get from your data collection. They represent the reality captured in your study—what really happened in your sample. In our problem, these are the numbers of vaccinated and non-vaccinated individuals belonging to different racial groups.
In a Chi-Square Test, observed frequencies are compared against expected frequencies to see if there are significant differences. For instance, in the exercise, '136 AAPI who completed vaccinations' and '1170 White who completed vaccinations' are examples of observed frequencies.
The core idea is to determine if the observed distribution of frequencies differs significantly from what was expected under the null hypothesis. By examining this difference, you can discern whether the association between the categories is statistically significant or if it's likely due to chance.The sum of the squared differences between observed and expected frequencies divided by the expected frequencies gives the Chi-Square statistic \\( X^2 = \sum \frac{(O-E)^2}{E} \). This measure helps statistically validate or refute the initial hypothesis assumptions.

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Most popular questions from this chapter

In a 2016 article published in the Journal of American College Health, Heller et al. surveyed a sample of students at an urban community college. Students' ages and frequency of alcohol use per month are recorded in the following table. Because some of the expected counts are less than 5, we should combine some groups. For this question, combine the frequencies \(10-29\) days and Every day into one group. Label this group \(10+\) days and show your new table. Then test the new table to see whether there is an association between age group and alcohol use using a significance level of \(0.05\). Assume this is a random sample of students from this college. $$ \begin{array}{lcccc} \hline & & \text { Alcohol Use } \\ \text { Age } & \text { None } & \text { 1-9 days } & \text { 10-29 days } & \text { Every day } \\ \hline 18-20 & 182 & 100 & 27 & 4 \\ \hline 21-24 & 142 & 109 & 35 & 4 \\ \hline 25-29 & 49 & 41 & 5 & 2 \\ \hline 30+ & 76 & 32 & 8 & 2 \end{array} $$

In July 2013 , Jeff Witmer obtained a data set from the Tampa Bay Times after the Zimmerman case was decided. Zimmerman shot and killed Trayvon Martin (an unarmed black teenager) and was acquitted. The data set concerns "stand your ground" cases with male defendants. Some of these were fatal attacks, and some were not fatal. Many of those charged used guns, but some used various kinds of knives or other methods. $$ \begin{array}{|lccc|} \hline & & \text { Accused } \\ \hline & \text { Nonwhite } & \text { White } & \text { All } \\ \hline \text { Not Convicted } & 48 & 80 & 128 \\ \hline \text { Convicted } & 19 & 38 & 57 \\ \hline \text { All } & 67 & 118 & 185 \\ \hline \end{array} $$ a. What percentage of the nonwhite defendants were convicted? b. What percentage of the white defendants were convicted? c. Test the hypothesis that conviction is independent of race at the \(0.05\) level. Assume you have a random sample.

Preschool Attendance and High School Graduation Rates for Males The Perry Preschool Project data presented in exercise \(10.39\) can be divided to see whether there are different effects for males and females. The table shows a summary of the data for males (Schweinhart et al. 2005). $$ \begin{array}{|lcc|} \hline & \text { Preschool } & \text { No Preschool } \\ \hline \text { HS Grad } & 16 & 21 \\ \hline \text { HS Grad No } & 16 & 18 \\ \hline \end{array} $$ a. Find the graduation rate for males who went to preschool, and compare it with the graduation rate for males who did not go to preschool. b. Test the hypothesis that preschool and graduation are associated, using a significance level of \(0.05\). c. Exercise \(10.40\) showed an association between preschool and graduation for just the females in this study. Write a sentence or two giving your advice to parents with preschool-eligible children about whether attending preschool is good for their children's future academic success, based on this data set.

Fill in the blank by choosing one of the options given: Chi-square goodness- of-fit data are often summarized with (one row or one column of observed counts-but not both, or at least two rows and at least two columns of observed counts).

The following table shows the average number of vehicles sold in the United States monthly (in millions) for the years 2001 through 2018 . Data on all monthly vehicle sales for these years were obtained and the average number per month was calculated. Would it be appropriate to do a chi-square analysis of this data set to see if vehicle sales are distributed equally among the months of the year? If so, do the analysis. If not, explain why it would be inappropriate to do so. (Source: www.fred.stlouisfed.org) $$ \begin{array}{|l|l|} \hline \text { Month } & \text { Avg Sales per Month (in millions) } \\ \hline \text { Jan } & 15.7 \\ \hline \text { Feb } & 15.7 \\ \hline \text { Mar } & 15.8 \\ \hline \text { Apr } & 15.8 \\ \hline \text { May } & 15.8 \\ \hline \text { June } & 15.7 \\ \hline \text { July } & 16.1 \\ \hline \text { Aug } & 16.1 \\ \hline \text { Sept } & 15.8 \\ \hline \text { Oct } & 15.9 \\ \hline \text { Nov } & 15.9 \\ \hline \text { Dec } & 15.9 \\ \hline \end{array} $$

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