00:01
For this problem, we are asked first to determine whether it is appropriate to use the normal distribution to estimate the p value for a test, using the sample results p hat equals 0 .16, with n equals 10.
00:12
Or, excuse me, not 10, n equals 100.
00:15
So to begin, we want n times p hat, which will be equal to 16 here, and n times q hat, or 1 minus p hat, which will be equal to 84.
00:26
Both of these are greater than 10, so this passes the test.
00:31
Now for actually doing the test, since we are at 5 % significance, we'll note that our critical z value here is going to be equal to 1 .645.
00:42
And additionally, we are doing a left -tailed test, which means that we will reject if our calculated z value is less than negative 1 .645...