Q. 5.3 - The Practice of Statistics for AP

Question

Sample surveys on sensitive issues can give different results depending on how the question is asked. A University of Wisconsin study randomly divided 2400 respondents into three groups. All participants were asked if they had ever used cocaine. One group of 800 was interviewed by phone; 21% said they had used cocaine. Another 800 people were asked the question in a one-on-one personal interview; 25% said “Yes.” The remaining 800 were allowed to make an anonymous written response; 28% said “Yes.”

3. Are the differences between the three groups statistically significant? Give appropriate evidence to support your answer. 

Step-by-Step Solution

Answer

Since $p-value \approx 0.001334>𝛼\left(=0.05\right)$, we can say that the difference is statistically significant.

Given

800 people were interviewed by phone; 21% said they had used cocaine. Another 800 people were asked the question in a one-on-one personal interview; 25% said “Yes.” The remaining 800 were allowed to make an anonymous written response; 28% said “Yes.” 

Table

The two way response table (observed) for the given data is:

Category	Yes	No
Direct	162	638
One on one	200	600
Anonymous	224	576

The two-way expected table for the given data is:

Category	Yes	No
Direct	195.3333	604.6667
One on one	195.3333	604.6667
Anonymous	195.3333	604.6667

Calculations

Since each of the frequencies are greater than 5, therefore we can apply ${𝓧}^2$ test.

We have,

${𝝌}^2=\sum _i\frac{{\left(O_i-E_i\right)}^2}{E_i}$

Upon substitution we get:

${𝝌}^2=\frac{{\left(162-195.3333\right)}^2}{195.3333}+\frac{{\left(200-195.3333\right)}^2}{195.3333}+\frac{{\left(224-195.3333\right)}^2}{195.3333}+\frac{{\left(638-604.6667\right)}^2}{604.6667}+\frac{{\left(600-604.6667\right)}^2}{604.6667}+\frac{{\left(576-604.6667\right)}^2}{604.6667}$

${𝝌}^2\approx 13.2395$.

Conclusion

The p-value can be calculated using the calculator is:

$p-value \approx 0.001334.$

Since $p-value \approx 0.001334>𝛼\left(=0.05\right)$, we reject $H_0$under the favor of $H_1$.