## Video transcript

In a local teaching district, a engineering award is available to teachers in order to install a bunch of four computers in their classroom. From the 6,250 teachers in the zone, 250 were randomly selected and asked if they felt that computers were an substantive teaching tool for their classroom. Of those selected, 142 teachers felt that the computers were an essential teaching instrument. And then they ask us, calculate a 99 % confidence time interval for the proportion of teachers who felt that the computers are an necessity teach cock. So let ‘s good think about the stallion population. We were n’t able to surveil all of them, but the stallion population, some of them fall in the bucket, and we ‘ll define that angstrom 1, they thought it was a good cock. They thought that the computers were a good creature. And we ‘ll just define a 0 value as a teacher that says not good. And some proportion of the total teachers think that it is a thoroughly learn tool. So that proportion is p. And then the rest of them think it ‘s a bad eruditeness tool, 1 minus p. We have a Bernoulli Distribution correct over here, and we know that the mean of this distribution or the have a bun in the oven value of this distribution is actually going to be p. So it ‘s actually going to be a value, it ‘s neither 0 or 1, then not an actual value that you could actually get out of a teacher if you were to ask them. They can not say something in between good and not good. The actual expected value is something in between. It is p. now what we do is we ‘re taking a sample of those 250 teachers, and we got that 142 felt that the computers were an essential education tool. then in our view, so we had 250 sampled, and we got 142 said that it is dear, and we’ll say that this is a 1. So we got 142 1 ‘s, or we sampled 1, 142 times from this distribution. And then the respite of the meter, so what ‘s left over ? There ‘s another 108 who said that it ‘s not good. therefore 108 said not good, or you could view them as you were sampling a 0, right ? 108 plus 142 is 250. So what is our sample think of here ? We have 1 times 142, plus 0 times 108 divided by our sum number of samples, divided by 250. It is adequate to 142 over 250. You could even view this as the sample proportion of teachers who thought that the computers were a good teaching creature. now let me get a calculator out to calculate this. So we have 142 divided by 250 is equal to 0.568. then our sample distribution proportion is 0.568. or 56.8 %, either one. indeed 0.568. now let ‘s besides figure out our sample distribution variation because we can use it late for building our confidence interval. Our sample variance hera — so let me draw a sample variation — we ‘re going to take the slant sum of the square differences from the base and separate by this subtraction 1. So we can get the best calculator of the on-key discrepancy. So it ‘s 1 times — no, it ‘s the other manner actually about — we have 142 samples that were 1 minus 0.568 away from our sample mean, or we ‘re this far from the sample hateful 142 times, and we ‘re going to square those distances. Plus the other 108 times we got a 0, so we were 0 minus 0.568 away from the sample distribution intend. And then we are going to divide that by the total number of samples minus 1. That minus 1 is our adjuster so that we do n’t underestimate. so 250 subtraction 1. Let ‘s get our calculator out again. And so we have 100 — we put a parentheses around everything — I have 142 times 1 minus 0.568 squared, plus 108 times 0 minus — and you could obviously do parts of this in your capitulum, but I ‘m just going to write the hale thing out — minus 0.568 squared, and then all of that divided by 250 minus 1 is 249. so our sample distribution variability is — well, I ‘ll just say 0.246. It is peer to — it is our sample variation — I ‘ll write it over here — our sample distribution variation is equal to 0.246. If you were to take the squarely settle of that our actual sample standard deviation is going to be, let ‘s take the feather root of that suffice right over there, and we get 0.496 is peer to 0. I ‘ll just round that up to 0.50. So that is our sample standard deviation. now this interval, let ‘s think of it this room, we are sampling from some sampling distribution of the sample mean. So it looks like this over hera, it looks that over there. And it has some mean, and sol the mean of the sampling distribution of the sample distribution beggarly is actually going to be the same thing as this mean over here — it ‘s going to be the same beggarly respect — which is the like thing as our population proportion. We ‘ve seen this multiple times. And the sampling distribution’s standard diversion, so the standard diversion of the sampling distribution, so we could view that as one standard deviation right over there. So the standard deviation of the sampling distribution, we ‘ve seen multiple times, is equal to the standard deviation — let me do this in a different color — is equal to the standard diversion of our original population divided by the squarely root of the issue of samples. So it ‘s divided by 250. now we do not know this right over here. We do not know the actual standard deviation in our population. But our best estimate of that, and that ‘s why we call it confident, we ‘re confident that the real number mean or the very population proportion, is going to be in this time interval. We ‘re convinced, but we ‘re not 100 % indisputable because we ‘re going to estimate this over here, and if we ‘re estimating this we ‘re actually estimating that over there. therefore if this can be estimated it’s going to be estimated by the sample standard deviation. so then we can say this is going to be approximately, or if we did n’t get a eldritch, completely skew sample, it actually might not even be approximately if we merely had a actually strange sample. But possibly we should write convinced that — we are convinced that the standard deviation of our sampling distribution is going to be approximately, rather of using this we can use our standard deviation of our sample distribution, our sample standard diversion. so 0.50 divided by the square root of 250, and what ‘s that going to be ? That is going to be — so we have this value right over here, and actually I do n’t have to round it, divided by the square ancestor of 250. We get 0.031. so this is equal to 0.031 over here. So that ‘s one criterion deviation. now they want a 99 % confidence interval. So the way I think about it is if I randomly pick a sample from the sampling distribution, what ‘s the 99 % chance, or how many — let me think of it this way. How many standard deviations away from the mean do we have to be that we can be 99 % convinced that any sample distribution from the sampling distribution will be in that interval ? so another way to think about think it, think about how many standard deviations we need to be away from the beggarly, so we ‘re going to be a certain number of standard deviations aside from the think of such that any sample distribution, any mean that we sample from here, any sample from this distribution has a 99 % chance of being plus or minus that many criterion deviations. So it might be from there to there. So that ‘s what we want. We want a 99 % chance that if we pick a sample from the sampling distribution of the sample distribution mean, it will be within this many standard deviations of the actual think of. And to figure that out let’s look at an actual Z-table. So we want 99 % confidence. so another way to think about it if we want 99 % confidence, if we barely look at the upper half right over here, that orange area should be 0.475, because if this is 0.475 then this other share ‘s going to be 0.475, and we will get to our — oh blue, we want to get to 99 %, so it ‘s not going to be 0.475. We ‘re going to have to go to 0.495 if we want 99 % confidence. So this area has to be 0.495 over here, because if that is, that over here will besides be. So that their sum will be 99 % of the sphere. now if this is 0.495, this value on the omega table right here will have to be 0.5, because all of this sphere, if you include all of this is going to be 0.5. So it ‘s going to be 0.5 plus 0.495. It ‘s going to be 0.995. Let me make indisputable I got that justly. 0.995. So let ‘s look at our Z-table. so where do we get 0.995. on our z table ? 0.995. is pretty conclude, good to have a little error, it will be mighty over here — this is 0.9951. so another way to think about it is 99 — then this value right hera gives us the whole accumulative area up to that, up to our mean. so if you look at the entire distribution like this, this is the mean right over here. This tells us that at 2.5 standard deviations above the mean, so this is 2.5 standard deviations above the average. So this is 2.5 times the criterion deviation of the sampling distribution. If you look at this whole area, this hale area over here, if you look at the Z-table, is going to be 0.9951, which tells us that just this sphere correct over here is going to be 0.4951, which tells us that this area plus the symmetrical sphere of that many standard deviations below the think of, if you combine them, 0.4951 times 2 gets us to 99.2. then this unharmed area justly here is 99.992. so if we look at the sphere 2.5 standard deviations above and below the average — oh, let me be careful. This is n’t just 2.5, we have to add another digit of preciseness. This is 2.5, and the next finger of preciseness is given by this column over here. So we have to look all the manner up into the second to the last column, and we have to add a finger of 8 here. So this is 2.58 standard deviations. We have 2.5 over here, and then we get the following finger 8 from the column. 2.58 standard deviations above and below the standard deviation encompasses a fiddling over 99 % of the total probability. So there ‘s a little over a 99 % opportunity that any sample mean that I select from the sampling distribution of the sample entail will fall within this much of the standard diversion. So let me put it this way. There is a 99 — it ‘s actually, what, a 99.2 % gamble, right ? If you multiply this times 2 you get 0.99 — actually you get 0.9902. So we ‘ll say roughly 99 % chance that any sample that a random sample distribution mean is within 2.58 standard deviations of the sampling beggarly of the bastardly of the sampling distribution of the sampling mean, which is the same thing as our actual population entail, which is the like thing as our population symmetry. So of p. And we know what this value is good here. At least we have a becoming estimate for this respect. We do n’t know precisely what this is, but our best estimate for this value is this over here. So we could re-write this, so we could say that we are confident because we are actually using an calculator to get this value here. We are confident that there is a 99 % luck that a random ten, a random sample mean, is within — and let ‘s calculate out this measure right hera using a calculator. So it is 2.58 times our best calculate of the standard deviation of the sampling distribution, so time 0.031 is equal to 0.0 — well let’s fair round this up because it ‘s so close to 0.08 — is within 0.08 of the population proportion. Or you could say that you’re convinced that the population proportion is within 0.08 of your sample distribution beggarly. That ‘s the demand same affirmation. so if we want our confidence interval, our actual issue that we got for there, our actual sample hateful we got was 0.568. So we could replace this, and actually let me do it. I can delete this right here. Let me gain it. I can replace this, because we actually did take a sample distribution. So I can replace this with 0.568. So we could be convinced that there ‘s a 99 % find that 0.568 is within 0.08 of the population proportion, which is the like thing as the population mean, which is the same thing as the average of the sampling distribution of the sample entail, sol away and so on. And just to make it clear we can actually swap these two. It would n’t change the mean. If this is within 0.08 of that, then that is within 0.08 of this. So let me switch this up a little bite. So we could put a phosphorus is within of — let me switch this up — of 0.568. And now linguistically it sounds a fiddling snatch more like a confidence interval. We are confident that there ‘s a 99 % casual that p is within 0.08 of the sample distribution mean of 0.568. So what would be our confidence interval ? It will be 0.568 plus or minus 0.08. And what would that be ? If you add 0.08 to this justly over here, at the upper end you ‘re going to have 0.648. And at the lower end of our scope, so this is the upper end, the lower conclusion. If we subtract 8 from this we get 0.488. So we are 99 % confident that the on-key population proportion is between these two numbers. Or another way, that the on-key percentage of teachers who think those computers are well ideas is between — we ‘re 99 % confident — we ‘re confident that there ‘s a 99 % gamble that the truthful share of teachers that like the computers is between 48.8 % and 64.8 %. nowadays we answered the first part of the question. The second part, how could the sketch be changed to narrow the confidence interval, but to maintain the 99 % confidence interval ? Well, you could barely take more samples. If you take more samples than our calculate of the standard deviation of this distribution will go down because this denominator will be higher. If the denominator is higher then this wholly thing will go down. so if the standard deviations go down hera, then when we count the standard deviations, when we do the asset or subtraction on the range, this measure will go down and will narrow our range. So you just take more samples.

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