Assume that a sample is used to estimate a population...

Assume that a sample is used to estimate a population proportion p. Find the 90% confidence interval for a sample of size 386 with 290 successes. Enter your answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places. < p < Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places.

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00:01 Okay, in this problem we have to assume that a sample is used to estimate the population proportion p.

00:05 Well, so we're saying the population proportion, but we have p hat.

00:09 And p hat is going to be the sample size of 386, so n is equal to 386, and then our x is equal to 290.

00:18 That's how many successes we have.

00:20 So p hat is going to be 290 divided by 386.

00:23 290 divided by 386, and that's going to give us 0 .75129.

00:29 So 0 .751295.

00:34 That's what i'm saying, 95.

00:37 You enter your answer as a trilinear inequality, so we need a 90 percent confidence interval for this.

00:46 So what we're going to do is we're going to say that we have p hat plus or minus the z -score that's associated with 90 percent confidence, and then we're going to multiply that times the square root of p hat times 1 minus p hat, and then over n.

01:05 Okay, so the z -score that's associated with 90 percent is 1 .65.

01:12 So we have 0 .751.

01:14 I'm just going to round it to 3 .7.

01:17 Come on now, 0 .751 plus or minus, uh, what was it again? i just said 1 .65.

01:25 Uh, 1 .65.

01:27 Actually, it wants it to three decimal places, doesn't it? well, we'll let our calculator worry about that.

01:32 We'll only round it at the very end...