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In the study referenced in exercise \(10.33\), researchers also collected data on use of apps to monitor diet and calorie intake. The data are reported in the table. Test the hypothesis that diet app use and gender are associated. Use a \(0.05\) significance level. $$ \begin{array}{ccc} \text { Use } & \text { Male } & \text { Female } \\ \hline \text { Yes } & 43 & 241 \\ \hline \text { No } & 50 & 84 \\ \hline \end{array} $$

Short Answer

Expert verified
We reject the null hypothesis and conclude that there is a significant association between gender and diet app usage based on the Chi-Square test with a p-value virtually 0, which is less than the significance level 0.05.

Step by step solution

01

Define Null and Alternative Hypothesis

The null hypothesis (H0) is that using a diet app and the gender of a person are independent. The alternative hypothesis (H1) is that using a diet app and the gender of a person are not independent.
02

Compute Expected Frequencies

To perform the Chi-Square test, we first have to calculate the expected frequencies for each cell in the table. The expected frequency is calculated by multiplying the row total and the column total of each cell and dividing by the total population. In this case, expected frequencies for 'Male who use app' = (43+50)*(43+241) / 418 = 94.54, 'Male who don't use app' = (43+50)*(50+84) / 418 = 50.46, 'Female who use app' = (241+84)*(43+241) / 418 = 189.46, 'Female who don't use app' = (241+84)*(50+84) / 418 = 84.54
03

Calculate Chi-Square Statistic

The chi-square statistic is calculated by summing the squared difference between observed and expected frequencies divided by the expected frequency, for all cells. Chi-Square statistic = ((43-94.54)^2/94.54)+((50-50.46)^2/50.46)+((241-189.46)^2/189.46)+((84-84.54)^2/84.54) = 80.85
04

Find Chi-Square Critical Value and P-value

With a degree of freedom = (2-1)*(2-1) =1, and significance level 0.05, the critical chi-square value from the chi-square distribution table is 3.841. P-value associated with a Chi-square value of 80.85 is virtually 0.
05

Decision

Since the chi-square statistic is greater than the critical value, we reject the null hypothesis. Also, the P-value is less than 0.05, which gives further evidence to reject the null hypothesis.

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Key Concepts

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

Null Hypothesis
Understanding the null hypothesis is crucial when conducting any statistical test, including the Chi-Square Test for Independence. The null hypothesis, often denoted as \(H_0\), is a statement that there is no effect or no relationship between two categorical variables. In the context of a Chi-Square Test, it suggests that the variables are independent, meaning the occurrence of one does not affect the probability of the occurrence of the other.

For example, when evaluating if gender is associated with the use of diet apps, the null hypothesis would be that gender does not influence a person’s likelihood to use diet apps. This is a starting point for statistical tests and helps researchers determine whether their observed data can be explained by chance or if there is statistical evidence of an association.
Expected Frequencies Calculation
The Expected Frequencies Calculation is a pivotal step in conducting a Chi-Square Test. It helps ascertain what the data would look like if the null hypothesis were true. To calculate the expected frequencies, we multiply the total for each row by the total for each column and then divide by the grand total of all observations.

Formula for Expected Frequency

This calculation is based on the formula: \(E_{ij} = \frac{(\text{row total}_i) \times (\text{column total}_j)}{\text{grand total}}\), where \(E_{ij}\) denotes the expected frequency for the cell at row \(i\) and column \(j\). By calculating these expected frequencies, we establish a benchmark against which the observed frequencies will be compared in the subsequent steps of the Chi-Square Test.
Chi-Square Statistic
The Chi-Square Statistic is a measure of how much the observed frequencies diverge from the expected frequencies. If the variables are truly independent, as the null hypothesis claims, the observed data should closely match the expected data, and the Chi-Square statistic should be small. A large Chi-Square statistic indicates a higher likelihood of a significant association between the variables.

To calculate this statistic, we use the formula: \(\chi^2 = \sum {\frac{(O - E)^2}{E}}\), where \(^2\) is the Chi-Square statistic, \(O\) is the observed frequency, and \(E\) is the expected frequency. For each cell in the table, we subtract the expected frequency from the observed frequency, square this difference, and divide by the expected frequency. Summing these values provides the overall Chi-Square statistic.
P-value Analysis
The P-value is a probability that measures the evidence against the null hypothesis. A P-value analysis helps us determine the significance of the results. After calculating the Chi-Square statistic, we use it to find the P-value, which tells us the probability of observing a Chi-Square statistic as extreme as, or more extreme than, what we calculated if the null hypothesis were true.

If the P-value is less than the chosen significance level (commonly \(\alpha = 0.05\)), we reject the null hypothesis. This indicates that the observed data is too unlikely to have occurred by random chance, given the assumption that the null hypothesis is true, and suggests a statistically significant association between the variables. In our example, a very small P-value indicates strong evidence to reject the null hypothesis, implying a likely association between diet app use and gender.

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