00:01
So for this problem, we effectively have that x, the number of people who continued to stop smoking.
00:10
I'll just say number of quitters for short, is effectively going to be distributed as a binomial random variable, where we are going to be going under the assumption that, well, we're doing this for 498 trials, 498 trials, and we're effectively assuming that the probability of somebody abstaining from smoking is going to be 0 .06.
00:36
So we have that our actual value of our test statistic here, x, is actually just directly going to be that 47 number who stopped smoking.
00:48
The p value is going to be the probability x is greater than or equal to 47 given that x is distributed as a binomial distribution.
01:02
So we would have that that would be more easily calculated as probability that x or 1 minus the probability that x is less than or equal to 46, which would then in turn be 1 minus the sum from i equals 0 up to 46 of 498 choose i times 0 .06 0 .06 hour of i times 0 .94 hour of 498 minus i now i'm going to pause and calculate that all right so i find that the result here is probability of 0 .0016 roughly roughly which because this is a very very small p value we can conclude that the incentive appears to work incentive works at for instance the alpha equals 0 .01 or p less than 0 .01 level then using the okay so moving on we're asked to find the mean and standard error of the normal distribution that most closely matches this so the idea here is that we can approximate a binomial by a normal distribution where i'll call our normal random variable y.
02:37
The mean value of the normal distribution is going to be equal to the number of trials times the probability of success.
02:43
The standard deviation of the normal distribution is going to be the square root of n times p times 1 minus p.
02:51
So, we have that the mean value, n times p, would be equal to 0 .06 times 498.
03:02
So, approximating this with a mean value of 29 .88 for the mean, and square root of 0 .6, or actually, i'll put it this way, it would be 0 .94 times 0 .96 times 4 .98.
03:21
So we have a standard deviation equal to 5 .297.
03:29
So having that then, pardon me, i had to pause my recording for a moment there.
03:36
Now, this off.
03:39
So the p value in this new distribution would be the probability that y is greater than or equal to it was 47 given the information of that distribution.
03:52
And i'll find that just directly using my software here.
03:58
So that's one minus the cumulative distribution function for a normal distribution with mean value 29 .88, standard deviation, 5 .2997.
04:10
And we want to subtract off the area to the left of 47...