On this video we will see how to assess the Z-teststatistic and p-worth for the two percentage Z-test. The instructional materials for theTI-83 and 84 are same. And we are going to with this instance, the equal one we used for thetwo percentage Z-interval. Simply to summarize the quandary,we have now a current provider and now we have a potential supplier. And we wish to understand is there proof that the prospective provider has a better expense of passing the inspection considering the fact that if they do we’regoing to move with them. So our null hypothesis isthat the real proportions that will pass inspection arethe same for the two companies and our alternate hypothesis, the one we’re looking to seeif there is evidence for, is that the prospectivesupplier has a greater expense of passing inspectionthan the present provider. So, first we carry out the preliminary steps. We set our significance degree alpha, we need to anticipate that there may be two unbiased,random samples right here.And now, specific from when we did the two proportionZ-interval, we’re going to look at this variety, which isthe pooled percentage, p-hat. And the rationale for this isbecause we have now a null declare that says that the p1 equals p2. So in a similar way to after we didthe one proportion Z-test and we had a hypothesized price for p, we use that hypothesized valuein checking our conditions and in calculating the SE. So here what we’re going todo is use our fine estimate for what this percentage thatthese two equal maybe. And so our quality estimate for that is to mix our two sample proportions. And we try this by addingthe whole number of yeses over the whole pattern size and here we get 0.9285. So this quantity we will use once we verify our conditionsand we’re also going to use it in the SE method. So right here we now have ourn1[p-hat], n1[1 minus p-hat] n2[p-hat] and n2[1 minus p-hat] are all better than or equal to 10.And now when we calculate our SE, we must go to the formulation sheet and we’re looking at thedifference of sample proportions and we’ve this specialcase where p1 equals p2. This is our null speculation. So for the reason that we now have this hypothesis right here, we’re going to use thisstructure written right here for our SE. So we’ve got p-hat right here, 1minus p-hat, et cetera. So we fill within the numbers, and now we’re equipped tograb the calculator. So, we perpetually start with STAT, exams, and this time we’re goingto go to 2-PropZTest. So let’s go to STAT, exams,and find 2-PropZTest. 2-PropZTest, here it’s. So x1 is 899, n1 is thesample dimension of one thousand, x2 is 958, do not forget x1 andx2 have got to be integers. Of course, n1 and n2 haveto be integers as good. So n2 is also a thousand.And now our alternate claimis that p1 is lower than p2, so we’ll pick the less than and now we’ll do Calculate. And we see we get a Z-ranking of -5.12 and a p-value very, very small. Become aware of this also gives youp-hat, the pooled share. So here is our .9285, whichis what we calculated here they usually healthy. So what’s our conclusion? Our p-worth is far less than alpha, so we reject H sub zero,now we have evidence that the potential provider hasa better true cross inspection than the present provider, so we will have to go together with theprospective provider.That is it for this video, when you liked it provide it athumbs up and subscribe beneath. Thanks for observing..