Determine whether a normal sampling distribution can be used...

Determine whether a normal sampling distribution can be used for the following sample statistics. If it can be used, test the claim about the difference between two population proportions $p_1$ and $p_2$ at the level of significance $\alpha$. Assume that the samples are random and independent. Claim: $p_1 \neq p_2$, $\alpha = 0.01$ Sample Statistics: $x_1 = 32$, $n_1 = 68$, $x_2 = 33$, $n_2 = 58$ Determine whether a normal sampling distribution can be used. The samples are random and independent. A normal sampling distribution (Round to two decimal places as needed.) be used because $n_1\hat{p} = \square$, $n_1\hat{q} = \square$, $n_2\hat{p} = \square$, and $n_2\hat{q} = \square$

Transcript

00:01 Alright, so we are looking at whether or not two proportions are equal or not.

00:05 So that's our pair of hypotheses.

00:09 Our null is that p1 is equal to p2.

00:12 The alternative is that p1 is not equal.

00:21 See if they're different.

00:23 And we're going to test this at the alpha of 0 .01 level of significance, which means we would reject the null hypothesis if the p -value that we find is less than this, alpha of 0 .01.

00:37 So we're given the data here.

00:39 Population 1, there was a sample of 72.

00:43 37 met some criteria.

00:45 Population 2, 62 were sampled and 36 met that same criteria.

00:50 And well, the first question is, can we determine whether or not we can use a normal sampling distribution? the reason is yes, we can, because the way you check for that is if n times p is greater than 5 and n times the complement, 1 minus p, is greater than 5.

01:15 You might even sometimes see this as 10, 5 or 10, depending on how strict you want to be, but a common one that we see is 5.

01:28 And what that means is that basically you have at least 5 in each of the two categories, right? because 37 out of 72 met this criteria.

01:40 Well, how many did not meet the criteria? more than 5, right? because 37 from 72 is a number greater than 5 and same can be said for population 2.

01:47 So yes, we can use normal sampling distribution here.

01:51 All right, and that's going to be the z standard normal distribution.

01:57 So the formula for a difference of proportions is the following.

02:00 P hat 1 minus p hat 2 divided by the square root of p times 1 minus p multiplied by 1 over n1 plus 1 over n2.

02:13 Now this n1, n2, those are the sample sizes from the two groups, 72 and 62 respectively for 1 and 2.

02:21 And this p here, this p is not the p that we're comparing our alpha to.

02:26 This p is actually the pooled proportion.

02:30 Pooled proportion.

02:31 And the way we get that is by taking the sum of each groups, the sum of the people who met the criteria or whatever the items met the criteria, elements, divided by the sum of all the elements sampled or all the people sampled, whatever it is you're looking at.

02:55 So add 37 and 36, that's your numerator.

02:58 Add 72 and 62, that's our denominator.

03:01 There we go.

03:02 And then, so 73 out of 134.

03:04 These are the proportions.

03:07 P hat 1 is .513 or .514.

03:14 Group p hat 2 is .581.

03:18 And then the pooled is .545...

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