WEBVTT

1
00:00:00.000 --> 00:00:09.961
Here is the problem: A sample of 200 has a sample proportion of 0.42. Find the 99 percent confidence interval for the proportion.

2
00:00:10.411 --> 00:00:15.136
Build a 99 percent confidence interval for the population proportion p.

3
00:00:15.936 --> 00:00:17.213
Sample proportion.

4
00:00:17.663 --> 00:00:18.673
Standard error.

5
00:00:19.123 --> 00:00:36.352
Critical value for 99 percent confidence. The middle 99 percent leaves 0.5 percent in each tail. Large-sample condition met: np̂ approximately 84 and n(1 minus p̂) approximately 116 are both at least 10.

6
00:00:36.802 --> 00:00:37.858
Margin of error.

7
00:00:38.659 --> 00:00:41.608
Interval: estimate plus or minus margin.

8
00:00:42.408 --> 00:00:56.491
So the answer is 99 percent CI: (0.330104, 0.509896); margin of error 0.0898961.

9
00:00:56.941 --> 00:01:18.083
We are 99 percent confident that the population proportion p lies between 0.330104 and 0.509896. The confidence describes the method: about 99 percent of intervals built this way from repeated samples would capture the true value.

10
00:01:18.883 --> 00:01:30.249
The interval is centred on the estimate: (0.330104 plus 0.509896) / 2 equals 0.42.

11
00:01:30.699 --> 00:01:39.012
Large-sample condition met: np̂ approximately 84 and n(1 minus p̂) approximately 116 are both at least 10.