A survey of adults in the United States found that 61% ate at ...

Question

A survey of adults in the United States found that 61% ate at a restaurant at least once in the past week. You randomly select 30 adults and ask them whether they ate at a restaurant at least once in the past week. (Source: Gallup)

c. Is it unusual for exactly 14 out of 30 adults to have eaten in a restaurant at least once in the past week? Explain your reasoning.

c. Is it unusual for exactly 14 out of 30 adults to have eaten in a restaurant at least once in the past week? Explain your reasoning.

Accepted Answer

Hi, everyone, let's take a look at this practice problem. This problem says a recent poll found that 54% of college students drink coffee daily. You randomly select 40 students and ask if they drink coffee every day. Is it unusual for exactly 18 out of 40 students to report drinking coffee daily? Select the appropriate reason, and we're given 4 possible choices as our answers. For choice A, we have yes, because 18 is much lower than the expected value. For choice B, yes, because the probability of getting exactly 18 is less than 5%. For choice C, we have no, because the probability of getting exactly 18 successes is greater than 5%, and for choice D, we have no, because the sample size is too small to make a conclusion. So, to determine if exactly 18 out of 40 students reporting daily coffee drinking is unusual, we evaluate how likely this outcome is based on a binomial distribution. So, for our distribution, we're gonna have a proportion, P that is going to be equal to 0.54, which is our 54%, once we convert that percentage into a decimal. We also have our sample size, which is going to be in, and that's gonna be equal to 40. And we're going to have our observed count X, which is going to be equal to 18. Now, we'd like to basically approximate our binomial distribution with a normal distribution. And to check to see if we can do that, we'll need to calculate two quantities. First, quantity we'll need to calculate is going to be N multiplied by P, which is going to be equal to. 40, multiplied by 0.54. And when we evaluate this expression, this is going to be equal to 21.6. The other quantity that we need to calculate is N, multiplied by the quantity of 1 minus P, which is going to be equal to 40, multiplied by the quantity of 1 minus 0.54. In quantity, and when we evaluate this expression, this is going to be equal to. 18 0.4. Now, since both of these quantities are greater than are equal to 5, that means that we can actually approximate our binomial distribution with a normal distribution. Next, we'll need to calculate our Z score, and for that we're gonna need to calculate our mean, which is going to be mu. And recall that that is going to be equal to N multiplied by P for binomial distribution, and we've actually already calculated this quantity, and that was equal to 21.6. The other quantity that we need is our standard deviation, sigma, and we call it for binomial distribution. This is going to be equal to the square root of the quantity of N multiplied by P, multiplied by the quantity of 1 minus P in quantity. And so substituting in our values, we see that this is going to be equal to the square root of the quantity of 40, multiplied by 0.54. Multiplied by the quantity of 1 minus 0.54. In quantity And when we evaluate this expression, this is going to be equal to the square root. Of 9.9. 36. Now, we also need to apply a continuity correction for X equal to 18. So, we're going to use the interval of 17.5 to 18.5. So, we basically subtracted 0.5 from 18, as well as added 0.5 to 18 to get our two values. So that means we're gonna need to calculate to Z scores. One, for an X value. That is equal to 17.5, we'll call that X1, and for other X value, we'll call it X2, that's going to be equal to 18.5. So, we call your formula for calculating the Z score, that's going to be Z is equal to the quantity of X minus mu in quantity divided by sigma. So, for our first C score, which we'll call Z1, this is going to be equal to X1, which is 17.5 minus U, which we calculated as 21.6, and then that quantity is going to be divided by our standard deviation, sigma, which is going to be the square root. of 9.936. And when we evaluate this expression, this is going to be approximately equal to. -1.301. And for other Z value, we'll call it Z2. This is going to be equal to X2, which is 18.5 minus mu, which is 21.6. And then that quantity is going to be divided by. Sigma, which is the square root of 9.936. And when we evaluate this expression, this is going to be approximately equal to. -0.983. So now we need to find the cumulative probabilities corresponding to these two Z scores in our standard normal distribution tables. And if we look for Z1, we see that it falls between two of our Z values in a table. It falls between Z equal to minus 1.30, which has a cumulative probability P of Z less than minus 1.30, that is equal to. 0.0968. And the other Z value is going to be Z equal to minus 1.31, which has a cumulative probability P of Z less than minus 1.31, that is equal to. 0.0951. And so since our Z value falls in between two Z values on our table, we'll need to use interpolation to find its cumulative probability. So the cumulative probability that we're looking for, which is going to be P of Z less than minus 1.301, is going to be equal to, and here we call your interpolation formula. So, the first term is going to be the change in our probability divided by our change in Z scores from our table. So, we'll have the quantity of 0.0968 minus 0.0951, and that quantity is going to be divided by the quantity of minus 1.30 minus. -1.31 in quantity, and then that fraction is going to be multiplied by the quantity of the Z score that we're looking for, which is -1.301. Minus and here we'll take the smaller of our two Z values, which is going to be -1.31. In quantity, and then we'll have to add to this, the probability for Z equal to -1.31. Which is 0.0951. And so, when we evaluate with this expression, this is going to be approximately equal to. 0.0966. So, that is our cumulative probability for Z1. Now, we do the same thing for Z2, and again, when we look up our Z value in our table, we see that falls in between two different Z values. It falls in between Z equal to. -0.98. Which has a cumulative probability of the less than. -0.98. That is equal to 0.1635. And the other Z value is going to be Z equal to minus 0.99, and this has a cumulative probability P of Z less than minus 0.99. And that's gonna be equal to. 0.1611. And so again, we'll have to use interpolation to find our cumulative probability. So, we'll be looking for the cumulative probability E of Z. Less than -0.983. And this is going to be equal to, and again, here we have to use the same interpolation formula. So we'll have the quantity, and the numerator here we have our change in our cumulative probabilities. So we'll have 0.1635 minus 0.1611, and that quantity is going to be divided by the change in our Z scores, which is going to be -0.98 minus 0.99. In quantity, that fraction is going to get multiplied by the quantity of minus 0.983, minus. -0.99 in quantity. And they'll have to add to that. The 0.1611. And so when we evaluate this expression, this is going to be approximately equal to. 0.1628. So, we're now going to use these cumulative probabilities to find the probability of getting exactly 18 students. So that probability P of 17.5 is less than X, which is less than 18.5. Using our Continuity correction. But this probability is going to be equal to the probability P of -1.301 is less than Z, which is less than -0.983. But that probability is going to be equal to the probability that P. Of Z less than minus 0.983. Minus the probability P of Z less than -1.301. And we actually have values for those two probabilities, so substituting them in. This is going to be equal to. 0.1628. Minus 0.0966. And when we evaluate that expression, that's going to be equal to. 0.0662. Well, we actually want that as a percentage, so we'll multiply that by 100%, and this is going to be equal to 6.62%. And we call that a Outcome is unusual if its probability is less than 5%. So, in our case here, our 6.62% is actually greater than 5% So that means the answer to this problem is going to be no. That's because our calculated probability of getting exactly 18 students as our outcome is greater than 5%. So if we take our answer here and compare it to our answer choices, we see that the correct answer choice actually corresponds to answer choice C. So, I hope that this explanation has been useful, and I'll see you in the next video.

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Key Concepts

Binomial Distribution

Normal Approximation

Unusual Events

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