In 2008 the better business bureau settled 75 of complaints...

In 2008 the Better Business Bureau settled 75$\%$ of complaints it received $(U S A$ Today, March $2,2009 ) .$ Suppose you have been hired by the Better Business Bureau to investigate the complaints it received this year involving new car dealers. You plan to select a sample of new car dealer complaints to estimate the proportion of complaints the Better Business Bureau is able to settle. Assume the population proportion of complaints settled for new car dealers is $.75,$ the same as the overall proportion of complaints settled-in $2008 .$ a. Suppose you select a sample of 450 complaints involving new car dealers. Show the sampling distribution of $p .$ b. Based upon a sample of 450 complaints, what is the probability that the sample pro- portion will be within .04 of the population proportion? c. Suppose you select a sample of 200 complaints involving new car dealers. Show the sampling distribution of $\overline{p} .$ d. Based upon the smaller sample of only 200 complaints, what is the probability that the sample proportion will be within .04 of the population proportion? e. As measured by the increase in probability, how much do you gain in precision by taking the larger sample in part (b)?

Transcript

00:01 Good morning, catherine t.

00:04 I'm going to help you answer this question about some customer complaints.

00:10 So you can see here we have the question prompt, which i'm sure you've already read.

00:16 I'm going to go right into answering the question.

00:20 So for part a, it says, suppose you select a random sample 450 complaints involving new car dealers show the sampling probability, sampling distribution of p.

00:35 So it really should say the sampling distribution of p hat.

00:39 When you have a true sampling distribution, there are some rules that you can follow.

00:46 And when we talk about the sampling distribution of p hat, we need to talk about the mean, the standard deviation, and the shape of the distribution.

00:56 So we're going to say the sampling distribution of p hat is equal to the sample proportion of complaints settled in a random sample of 400.

01:03 50 complaints and it has a mean mu of p hat equal to 0 .75.

01:08 So the sampling distribution of p hat will have the same mean as the population proportion.

01:13 So mu of p hat equals p.

01:15 So mu of p hat equals 0 .75 and a standard deviation of sigma p hat equals 0 .0 .04.

01:24 And you can see this is the formula here.

01:26 It's the square root of p times 1 minus p over n.

01:29 And when you do that math, you get 0 .02.

01:32 So make sure you show your work for this.

01:34 It's important.

01:36 And we can say the sampling distribution is approximately normal since our sample size times p, our mu of p hat .75, is equal to 337 .5, and you want to always round that up if you get a decimal.

01:49 So that's going to be 338.

01:51 And 450 our sample size times 1 minus mu of p hat .75 equals 1 .12 .5.

01:59 We're going to round that up since it's a decimal to 113, and they are both greater than or equal.

02:03 To 10.

02:05 So when you're describing a sampling distribution, you should talk about the mean of the sampling distribution, the standard deviation of the sampling distribution, and the shape.

02:14 So we can say it's approximately normal since n times p and n times 1 minus p are both greater than are equal to 10.

02:20 We could say it's approximately normal.

02:22 Now we like when things are normal, then we're able to use the normal cdf function in our calculator to find probabilities.

02:29 And that's what part b says.

02:31 So part b said based upon a sample of 450 complaints, what is the probability the sample proportion will be within 0 .04 of the population proportion.

02:41 Well, the population proportion is 0 .75.

02:45 So if i take 0 .75 and subtract 0 .04, i get 0 .71.

02:51 And if i take 0 .75 and add 0 .04, i get 0 .79.

02:55 So they want to know what's the probability that p hat will be within .71 and .79.

03:02 So we have our probability statement set up here.

03:05 We're going to use the normal cdf function in our calculator.

03:10 Your normal cdf function is for a lower bound and upper bound and the mean and standard deviation, which we just talked about in part a, the mean of the sampling distribution and the standard deviation of the sampling distribution...

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