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Breakfast Habits (Example \(1 \&\) 2) In a 2015 study by Nanney et al. and published in the Journal of American College Health. a random sample of community college students was asked whether they ate breakfast 3 or more times weekly. The data are reported by gender in the table. $$ \begin{array}{|lcc|} \hline \text { Eat breakfast at least } 3 \times \text { weekly } & \text { Females } & \text { Males } \\ \hline \text { Yes } & 206 & 94 \\ \hline \text { No } & 92 & 49 \\ \hline \end{array} $$ a. Find the row, column, and grand totals, and prepare a table showing these values as well as the counts given. b. Find the percentage of students overall who eat breakfast at least three times weekly. Round off to one decimal place. c. Find the expected number who eat breakfast at least three times weekly for each gender. Round to two decimal places as needed. d. Find the expected number who did not eat breakfast at least three times weekly for each gender. Round to two decimal places as needed. e. Calculate the observed value of the chi-square statistic.

Short Answer

Expert verified
Row Totals: Females = 298, Males = 143; Column Totals: Yes = 300, No = 141; Grand Total = 441. Percentage of students who eat breakfast at least three times weekly = 68.0%. The expected number of females who eat breakfast at least three times weekly = 201.81; The expected number of males = 98.19. The expected number of females who did not eat breakfast is 96.19, and the expected number of males is 44.81. The chi-square statistic can be computed by using the mentioned formula with the appropriate observed and expected frequencies for each cell.

Step by step solution

01

Calculate the row, column and grand totals

To get the total for each row, simply add the number of females and males. Similarly, to get the total for each column, add the 'Yes' and 'No' responses. The grand total is the sum of all values or the sum of the row or column totals. Here, the total number of females is \(206+92=298\) and the total number of males is \(94+49=143\). The total number of 'Yes' responses is \(206+94=300\) and the total number of 'No' responses is \(92+49=141\). The grand total is \(298+143=441\).
02

Calculate the overall percentage of students who eat breakfast at least three times weekly

To find the percentage of students who eat breakfast at least three times weekly, divide the total 'Yes' responses by the grand total and multiply by 100. \((300/441)*100 = 68.0%.\)
03

Calculate the expected number who eat breakfast at least three times weekly for each gender

The expected number is calculated by multiplying the row total by the column total and dividing by the grand total. For females, \((298*300)/441 = 201.81\). For males, \((143*300)/441 = 98.19.\)
04

Calculate the expected number who did not eat breakfast at least three times weekly for each gender

Similarly, for females, \((298*141)/441 = 96.19\). For males, \((143*141)/441 = 44.81.\)
05

Calculate the observed value of the chi-square statistic

The chi-square statistic is calculated using the formula \(\chi^2 = \sum [(O-E)^2/E]\) where \(O\) is the observed frequency and \(E\) is the expected frequency. In this exercise, it will require computing the chi-square for each cell, i.e., for males and females who answered 'Yes' and 'No' respectively. For instance, for females who eat breakfast at least three times weekly, O=206 and E=201.81. Therefore, \(\chi^2\) for this cell would be \((206-201.81)^2 / 201.81\). Compute the \(\chi^2\) for each of the remaining cells in the same way and sum the obtained values to get the total chi-square statistic.

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Key Concepts

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

Expected Frequencies
When dealing with categorical data, such as in the example where we have students' breakfast habits, expected frequencies are a fundamental concept. They give us a benchmark against which to compare the observed data. In the case of our study, we want to understand how many students we would expect to eat breakfast based on our observations, if gender had no influence on breakfast habits. This is calculated mathematically using the formula
Expected frequency (E) = (Row Total × Column Total) / Grand Total.
For example, to calculate the expected frequency of females who eat breakfast at least three times weekly, we use the row total for females and the column total for those who answered 'Yes', then divide by the grand total of all respondents. The result provides an expected count, assuming the distribution of breakfast habits is independent of gender. This concept is crucial when determining if there is an actual association between two categorical variables, such as gender and eating breakfast in our exercise. When the expected frequencies are significantly different from the observed frequencies, it may suggest an association worth exploring further.
Statistical Significance
The term 'statistical significance' serves as the cornerstone of hypothesis testing. It helps researchers determine whether their findings are due to chance or if there is an underlying effect. In other words, it measures whether the observed data deviates sufficiently from what we would expect to happen in a random distribution of results.

When we calculate the chi-square statistic using the observed and expected frequencies, we obtain a value that can then be compared against a critical value from the chi-square distribution table. If our calculated value is higher than the table value (considering our degrees of freedom and desired confidence level), the difference is considered statistically significant. In layman's terms, this means the pattern we observed in our data is probably not a fluke, and there could be real factors at play influencing the variables we are studying.
Categorical Data Analysis
Categorical data analysis is a statistical method used to analyze data where the variables represent categories, such as 'male' and 'female' or 'yes' and 'no' like we find in our exercise. Instead of numerical data points, we deal with counts or frequencies of occurrences. This type of analysis often includes techniques like chi-square tests, which compare our observed counts to expected counts under the assumption of no effect or no association (the null hypothesis).

Chi-square tests are commonly used and they help determine if the distribution of categorical variables differs from expected distributions. The breakfast habits study is a classic example of using this analysis to understand if there's a link between gender and frequency of eating breakfast, by comparing the observed distribution of meal patterns against expected counts.
Hypothesis Testing
Hypothesis testing is essentially a method of making decisions using data. Whether it's deciding if a new medication works or if there's a significant difference between groups, hypothesis testing provides a structured way to make those determinations. At its core are two competing hypotheses: the null hypothesis, symbolized as H0, which typically states that there is no effect or no difference; and the alternative hypothesis, H1, which states that there is an effect or a difference.

In the context of our example, we state the null hypothesis as 'there is no association between gender and eating breakfast at least three times weekly.' Through the chi-square test, we calculate a statistic that measures how much our observed data diverge from what we would expect under H0. If the divergence is large enough to be unlikely due to random chance (usually determined via a p-value), then we reject the null hypothesis in favor of the alternative. This process enables researchers to understand more deeply if their interventions or observations reflect a true underlying pattern or if they are seemingly random occurrences.

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Most popular questions from this chapter

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.

A penny was spun on a hard, flat surface 50 times, and the result was 15 heads and 35 tails. Using a chisquare test for goodness of fit, test the hypothesis that the coin is biased, using a \(0.05\) level of significance.

According to a 2017 report, \(64 \%\) of college graduate in Illinois had student loans. Suppose a random sample of 80 college graduates in Illinois is selected and 48 of them had student loans. (Source: Lendedu.com) a. What is the observed frequency of college graduates in the sample who had student loans? b. What is the observed proportion of college graduates in the sample who had student loans? c. What is the expected number of college graduates in the sample to have student loans if \(64 \%\) is the correct rate? Do not round off.

The table shows the results of rolling a six-sided die 120 times. $$ \begin{array}{|c|c|} \hline \text { Outcome on Die } & \text { Frequency } \\ \hline 1 & 27 \\ \hline 2 & 20 \\ \hline 3 & 22 \\ \hline 4 & 23 \\ \hline 5 & 19 \\ \hline 6 & 9 \\ \hline \end{array} $$ Test the hypothesis that the die is not fair. A fair die should produce equal numbers of each outcome. Use the four-step procedure with a significance level of \(0.05\), and state your conclusion clearly.

The Perry Preschool Project was created in the early 1960 s by David Weikart in Ypsilanti, Michigan. One hundred twenty three African American children were randomly assigned to one of two groups: One group enrolled in the Perry Preschool, and one did not enroll. Follow-up studies were done for decades to answer the research question of whether attendance at preschool had an effect on high school graduation. The table shows whether the students graduated from regular high school or not. Students who received GEDs were counted as not graduating from high school. This table includes 121 of the original \(123 .\) This is a test of homogeneity, because the students were randomized into two distinct samples. (Schweinhart et al. 2005 ) $$ \begin{array}{|lcc|} \hline & \text { Preschool } & \text { No Preschool } \\ \hline \text { HS Grad } & 37 & 29 \\ \hline \text { No HS Grad } & 20 & 35 \\ \hline \end{array} $$ a. For those who attended preschool, the high school graduation rate was \(37 / 57\), or \(64.9 \%\). Find the high school graduation rate for those not attending preschool, and compare the two. Comment on what the rates show for these subjects. b. Are attendance at preschool and high school graduation associated? Use a \(0.05\) level of significance.

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