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Hello everyone.
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So let's do the statistics problem, but before i get started, i recommend that you do the question yourself and come back to see if you got it right or not.
00:09
So hopefully you've done the question of some ways we're kind of together.
00:12
So we have a bunch of data we're given the random sample mean and standard deviations regarding students in a statistic class conducting a survey to estimate the mean number of unit students at their college that are enrolled in like, right? so what we want to do is find the 95 % confidence interval and we're given some options.
00:33
So i like to write down the data that we're given.
00:37
So we're given that n is equal to 49.
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We're given our x bar, our average, which is 122.
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And then we're given our sigma, which is 1 .6, right? standard deviation.
00:52
So 1 .6 units.
00:56
So from that, we can find the 95 % components integral.
01:08
So you can find our z critical value.
01:12
So if we go to the t -table or even the c -table, we can find that when the confidence intervals is 95%, our z -critical value is 1 .9.
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And we can even do that by finding the alpha.
01:24
So you can do 1 .905, which is 0 .05.
01:28
And you can find a z critical value when that is the probability.
01:33
We're kind of worked backwards for using the z table.
01:35
So we have the z critical value being 1 .96.
01:38
And we can just basically plug in all our values since we already have them.
01:42
We have to open to plus minus 1 .96, 9 .16 all over square meter of 49.
01:50
And then we'll find that.
01:52
So 12 .12 plus 90 .4 .8.
01:57
And then we can write down our interval.
01:59
So that would just end up looking like.
02:02
If we minus the two values, we get our lower bound, which is about 11 .7...