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 = 
                          = 
                          = 
                           =
                           = 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.