0:00
All righty.
00:01
So we've got a claim to test.
00:05
And the claim is that the proportion of man who owned cats is larger than 50 % at the 0 .05 level of significance.
00:14
So the null and alternative hypotheses would be.
00:16
So i see these selections here that came through.
00:19
It kind of came all jumbled.
00:20
But you know what? it's okay.
00:20
We don't need that.
00:21
We can figure this out without these selections here.
00:24
So the null is kind of the opposite of the claim that p is equal to point.
00:34
The alternative is that p is greater than 0 ,0, but than 0 .5.
00:44
We know that because it is larger than.
00:46
And this is going to be a right -tailed test because if here's our given mean, our given proportion, we want to see is it fall on this right side, this upper side.
00:58
So this is a right -tailed test.
01:01
And so we're given the sample of 90 men and 48 -owned cats.
01:06
The p value is it ends up being 0 .26 but here's how we get that so we're going to it's a z a z test for a sample proportion so we get z i have p hat here that's some sometimes is what you'll you'll see it as and that's given as p hat minus p and p hat is the point um the point estimate you have your sample statistic then you divide it by p times one minus p all over the square of n.
01:39
And this whole thing is in the denominator, the whole square root and the denominator.
01:44
And that's sig prop.
01:48
That's my little shorthand for, you might see this as sigma subp.
01:54
And that means it's a standard error of the sample proportion.
02:00
But that's what that is.
02:02
And so the whole denominator, when you substitute in the p of 0 .5, we get this number.
02:08
But just to show it all in there, so we get 0 .53, well, 53 repeating, minus 0 .5 all over the square root of 0 .5 times .5 again, because 1 minus .5 is .5 all over 90.
02:26
And we get this, the z score, .632.
02:36
And then from there we do a little table lookup.
02:45
And i used my spreadsheet to generate this value, this 0 .736.
02:52
That's the area to the left of the score we just found.
02:56
But we don't want that.
02:57
We want the area to the right.
02:59
So we found this area here to be 0 .736...