🎉 Announcing Numerade's $26M Series A, led by IDG Capital!Read how Numerade will revolutionize STEM Learning



Numerade Educator



Problem 35 Hard Difficulty

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)?


a. See graph
b. 0.9500
c. See graph
d. 0.8098
e. 0.1402


You must be signed in to discuss.

Video Transcript

So in this question, we're gonna be taking look at complaints that are handled by the Better Business Bureau. And it tells us to assume that the Better Business Bureau handles 75% of the complaints for new, uh, for new car dealerships. So we're going to assume that our population perimeter is 0.75 and they're going to take out Ah, a survey of 450 people. So our question is, what is the sampling distribution then for this sample? So, um, we need to look at our sample size to make sure our shape is going to be normal. So we have n times p for 50 times 500.75 and that comes out to 3 37.53 37.5, which is greater than or equal to 10 on end times one minus p. So all of the rest of them, um, or 0.25 times for 50 and that comes out toe 1 12.5 which is also greater than or equal to 10. So we know that our sampling distribution is going to be approximately normal, so I'm gonna draw my normal curve over here And what? This is going to be a distribution of IHS samples. And can we draw that look cleaner? This is our distribution of our see apples, Okay. And so we need to know. Then this center and the center is are expected number are what we expect to happen on average, and that is the same thing as P. So we expect. For the most part, most of our surveys are going to return a 75% of the complaints result. But we know that's not gonna happen every time that there's going to be some spread that's associated with that and the formula for calculating that is a p Times one minus p over. And so in this case, it's 0.75 times 0.25 over 4 50 and that comes out to be 0.2 04 And so since this is approximately normal, I'm gonna expect that my distribution spans three standard deviations in each direction. So 30.75 let's just kind of around this 2.2 So this would be 0.77 0.79 my 81 and then going in the in the negative direction subtracting too. Standard 0.2 Um, the standard deviation. We're gonna do that three times. So this is our sampling distribution of the samples. Okay, so the second question says, Okay, plus or minus 4%. What's the probability of plus or minus 4%? So we're talking about if we come over here to this graph, if we're at 75% we want to go down 2.71 and up, 2.79. You could see him not perfectly symmetric there, but I want to know what this area is in here. That's what we're talking about with this plus or minus 0.4%. So we can use our calculator and the normal CDF function to do that. So are 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 0.2 And I'm going to give it that little bit of extra Ah decimal place. Just that I don't make it a little bit more accurate. And this comes out to be 95% So 95% of the time we expect them to resolve within 4% of the population portion of the complaints. So Part C then says, Well, what if you don't have a sample of 4 50? 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 of your spread is going to increase. And let's take a look at why that happens. Um, so again, we want to find the sampling distribution so shape, center and spread. The shape is still going to be approximately normal. And we can verify that real quick because and Times P, which in this case is going to be point 75 times 200 is, um 1 50 and end times one minus P, which is 0.25 times 200 is 50 right to under. It, uh, is 1 50 plus 50. Okay, so we know it's approximately normal. The center is going to stay the same because we're still dealing with that true proportion that we're assuming of 75%. And the spread, then of all of our samples formula doesn't change 0.75 times 0.25. But now we're gonna divide by 200. And since we're dividing by a smaller number, that's going to increase the that the quotient, the answer to our division problem. And then we'll take the square root and we get 0.306 So we can see that for this sampling distribution. Most of it's the same, but the, um, spreads a little bit bigger, so I'm going to draw it again. These are our samples, and we know that it's approximately normal. And our center is 0.75. And so let's call this 0.31 right point. Uh huh. Not 0.31 0.31 big difference between those two things. So again, we're going to go up three center deviations and 0.31 endpoint 75 is 750.781 than another point. Another 0.31 It's gonna be 0.812 and then another point. Oh, 31 This 0.843 And then if we did the same thing going down, we're gonna subtract 0.31 and this is 0.719 and then 0.688 Hard to squeeze that in in this little drawing and my next 0.31 again, forget 0.657 so we can see that it's much more spread out from what we had before. So for Part D, it's asking us again the same question. What is the probability of getting something that's plus or minus 4% from the mean? So are still going to, um, 71 and 79? Because that's what 0.75 minus point of Ford 0.75 plus up 104 is. But I know now that it's going to be a little bit closer because our data is more spread out, right? So in 0.79 is maybe here so we can see that this area that we're talking about inside here there's a little bit smaller. So when I calculate my normal CDF, uh, I'm expecting and never to be a little bit smaller because my standard deviation is larger. This small, the same spread of numbers doesn't take up a smudge space. So here's our normal CDF, and I'm still going from 0.71 2.79. My mean is still 0.75. But now my standard deviation is 0.31 on this comes out to be 0.81. Okay, so when we had a larger sample size, we're expecting a greater proportion of our samples to be within a 4% margin of their or within 4% of our proportion. And now we have decreased Are the number of proportions we expect to be within that 4%. So the last question is part E. And and what percent, Um, have we lost or what is the difference between those two? And that is really just a matter of subtraction. So we have lost about 14% of our, um, of our accuracy by using that smaller sample size.