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)?

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)?

Key Concepts

Effect of Sample Size on Precision

The sample size (n) plays a critical role in determining the standard error of the sampling distribution. A larger sample size results in a smaller standard error, thereby increasing the precision of the estimate and reducing the margin of error. This principle demonstrates why increasing the sample size yields more reliable and precise estimates of the population parameter.

Probability Calculation Using the Normal Distribution

When the sampling distribution of the sample proportion is approximately normal, probability calculations can be performed using z-scores. This involves converting the margin of error into a z-score and then using the standard normal distribution to determine the probability that the sample proportion falls within a specified range of the true proportion.

Standard Error

Standard error quantifies the amount of variability or dispersion in the sampling distribution of a statistic. For a sample proportion, it is computed as sqrt[p(1-p)/n] and indicates the precision of the sample proportion as an estimate of the population proportion. The smaller the standard error, the more concentrated the sample proportion distribution is around the true proportion.

Sampling Distribution of a Sample Proportion

This concept refers to the probability distribution of the sample proportion (p?) over many samples drawn from the same population. It has a mean equal to the true population proportion (p) and a standard deviation (standard error) calculated as the square root of [p(1?p)/n]. This distribution provides the basis for inference about the population proportion using sample data.

Central Limit Theorem

The Central Limit Theorem states that for sufficiently large sample sizes, the sampling distribution of the sample proportion (or any other sample statistic) approximates a normal distribution, regardless of the underlying population distribution. This underpins the use of z-scores and normal probability calculations in estimating probabilities and margins of error in proportion problems.

Transcript

00:01 So in this question, we're going to be taking a look at complaints that are handled by the better business bureau.

00:06 And it tells us to assume that the better business bureau handles 75 % of the complaints for new car dealerships.

00:16 So we're going to assume that our population perimeter is 0 .75.

00:21 And they're going to take out a survey of 450 people.

00:26 So our question is, what is the sampling distribution? than for this sample.

00:33 So we need to look at our sample size to make sure our shape is going to be normal.

00:41 So we have n times p, 450 times .75, and that comes out to 337 .5, 337 .5, which is greater than or equal to 10.

00:54 And n times 1 minus p, so all of the rest of them, or 0 .25 times 450, and that comes out to 112 .5, which is also greater than or equal to 10.

01:08 So we know that our sampling distribution is going to be approximately normal.

01:14 Okay, so i'm going to draw my normal curve over here.

01:22 And what this is going to be a distribution of is samples.

01:29 And let me draw that a little cleaner.

01:38 This is our distribution of our samples.

01:42 Okay, and so we need to know then the center.

01:46 And the center is our expected number, or what we expect to happen on average.

01:53 And that is the same thing as p.

01:55 So we expect, for the most part, most of our surveys are going to return a 75 % of the complaints result.

02:04 But we know that's not going to happen every time that there's going to be some spread that's associated with that.

02:09 And the formula for calculating that is p times 1 minus p over n.

02:18 So in this case, it's 0 .75 times 0 .25 over 450.

02:26 And that comes out to be 0 .0 to 0 .04.

02:32 And so since this is approximately normal, i'm going to expect that my distribution spans three standard deviations in each direction.

02:42 So 0 .75, let's just kind of round this to 0 .02.

02:48 So this would be 0 .77, 0 .79, 0 .81.

02:53 And then going in the negative direction, subtracting to the standard .02, the standard deviation.

03:04 We're going to do that three times.

03:07 So this is our sampling distribution.

03:11 Of the samples.

03:13 Okay, so the second question says, okay, plus or minus 4%, what's the probability of plus or minus 4%.

03:24 So we're talking about, if we come over here to this graph, if we're at 75%, we wanna go down to 0 .71 and up to 0 .79, you could see i'm not perfectly symmetric there, but i wanna know what this area is.

03:43 In here.

03:44 That's what we're talking about with this plus or minus 0 .04%.

03:50 So we can use our calculator and the normal cdf function to do that.

03:57 So our lower boundary is going to be 71, our upper boundary is going to be 79, the mean is 0 .75, and the standard deviation is 0 .02, and i'm going to give it that little bit of extra decimal place, just, i don't know, make it a little bit more accurate.

04:15 And this comes out to be 95%.

04:19 So 95 % of the time we expect them to resolve within 4 % of the population proportion of the complaints.

04:32 So part c then says, well, what if you don't have a sample of 450? what if you have a sample of 200? so one of the things that happens is when your sample size goes down, your standard deviation or your spread is going to increase.

04:54 And let's take a look at why that happens.

04:58 So again, we want to find the sampling distribution...

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