Question:
The question is incomplete. The value of n and the required calculation was not given. Below is the remaining part of the question.
In 1999, n = 334
In 2009, n = 843
1 Estimate the difference between the proportion in 1999 and 2009
2. Find the standard error of this difference.
3. Construct and interpret a 95% confidence interval to estimate the true change, explaining how our interpretation reflects whether the interval contains 0.
Answer:
1. Difference between the proportion = 0.21
2. Standard error = 0.030471
3. There is a significant decrease.
Step-by-step explanation:
Given data:
Year 1999 2009
Sample size n1 = 334 n2 = 843
Sample proportion p1= 0.422 p2 = 0.212
1. The difference between the proportion is calculated as;
Difference = p1 -p2
= 0.422 - 0.212
=0.21
2. Calculating the standard error using the formula;
Standard error = [tex]\sqrt{\frac{p1(1-p1)}{n1}+ \frac{p2(1-p2)}{n2} }[/tex]
= [tex]\sqrt{\frac{0.4422(1-0.422)}{334}+\frac{0.212(1-0.212)}{843} }[/tex]
= [tex]\sqrt{0.00073 + 0.000198}[/tex]
=[tex]\sqrt{0.000928}[/tex]
= 0.030471
Therefore, the standard error = 0.030471
3. Calculating 95% confidence interval using the formula;
95% confidence interval = p1-p2 ±Za (SE(p1-p2))
= 0.422 - 0.212 ± 1.96*0.030471
= 0.21 ±0.059722
= 0.1503, 0.2697
0.1503 ≤ (p1-p2) ≤ 0.2697
The 95% confidence interval is (0.1503, 0.2697). The value 0 is not included in the 95% confidence interval. Hence, we may conclude that the students of Wisconsin binge at least three times in the past two weeks in the year 1999 and 2009 has been decreased.
