Minting Quarters Listed below are weights (grams) of quarters ...

Question

Minting Quarters Listed below are weights (grams) of quarters minted after 1964 (based on Data Set 40 “Coin Weights” in Appendix B).

b. Specifications require that the quarters have a weight of 5.670 g. What does the confidence interval suggest about that specification?

Weights of quarters: 5.7790, 5.5928, 5.6486, 5.6661, 5.5491, 5.7239, 5.5591, 5.5864, 5.6872, 5.6274 grams.

Accepted Answer

Hello there. Today we're gonna solve the following practice prom together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. A group of archaeologists are analyzing the weight in units of grams of pennies minted after 2000 to assess the consistency of their production. The specifications require that each penny weigh exactly 2.500 g. The following are the weights of a sample of 8 pennies 2.520, 2.510, 2.490, 2.530, 2.510, 2.495, 2.505, 2.515 using the sample data. Construct a 99% confidence interval for the true mean weight of pennies minted after 2000. What does the confidence interval suggest about the weight specification of 2.500 g? Awesome. So it appears for this particular problem we're asked to solve for two separate answers. Our first answers we're asked to use our sample data to help us construct or to create a 99% confidence interval for the true mean weight of the pennies minted after the year 2000. And our second answer we're trying to solve for is what does the confidence interval value that we solve for suggest about the weight specification of 2.500 g. So now that we know what we're ultimately trying to solve for, our first step that we need to take in order to solve this particular problem is we need to calculate the mean of the sample. So let us call the mean X bar, and X bar, as we should recall, since the equation to help us solve for the mean is we need to add up all of our data points. That are given to us, the sample of 8 pennies, we need to add all of these 8 value, values together, and then we need to divide by 2. So we need to take 2.520 plus. 2.510 plus 2.490 plus 2.530 plus 2.510 plus 2.495 plus 2.505 plus 2.515. Fantastic. And then we need to divide everything by 2, and when we do that, we will find that our mean of the sample X bar is going to be simply equal to drumroll, 2.509375 g. And now, our second step is we need to calculate the deviations from the mean for each value, and then we need to square them. And when we do that, I've gone ahead in order to make things simple, and you could do the same thing as well. But I created a table to help organize my data that I determined. So what we're doing is we have our far left column which represents our value for X subscript I. And then our right column is parentheses X I minus X2, and our values for X subscript I are 8 different weights for the sample pennies, so there's 8 different values listed, and ultimately, the sum of Xi is equal to 2.509, and the sum of parentheses Xi minus X bar. To the power 2 is equal to 0.001171875. And when we plug in our known variables, we will receive this following values and it's we, I wrote our data table in the same order that our sample weights appear in the prom itself. So it's 2.520 going down to 2.515, and there are corresponding weights when you plug in your known variables. Fantastic. So in an effort to save time, I won't read off each individual value for you, but just note that we just plug in our different values for XI and we subtract the mean, and then we square that value, the difference, we square the difference, and that's how we get our value in our right column. So, with that said, we now need to note that getting the sum of the deviations we will get. 0.001171875 So that will mean that when we take. Our other data points, so when we take 0.0028 + 0.0039 + 00. So 0.0067 plus 0.0018 plus 0.0039 plus 0.0059 plus 0.0045 plus 000. 0033 will be approximately equal to, so when we add all these values together, it'll be approximately equal to 0.0328. Fantastic. And in order to get the variance, as we should recall, we need to take N minus 1, which is going to be equal to 7, and sigma squared. So I'll make a put a line to separate, and then we need to note that sigma squared, which is our standard deviation. Squared is going to be equal to 0.001171875. And once again, let's note that we're trying to solve for the variance. And then we need to divide our value of 0.001171875, and we need to divide this by 7. And when we do that, we will find that it is approximately equal to 0.0001674. Fantastic. So now we need to take the square root to get our standard deviation. So like I said, We determined our variants, and now we're finding this our standard deviation, which is sigma all by itself, but right now we have sigma squared, so we have to take the square root of our value that we just found. So we need to take sigma is equal to the square root of 0.0001674, which will be approximately equal to 0.01294 g. Fantastic. So now, our third step is we need to construct the confidence interval for the mean. So using a T distribution with N minus 1 is equal to 7 degrees of freedom for a 99% confidence interval, the critical value, which is going to be denoted as T subscript alpha divided by 2, will be approximately equal to 3.499. And we also need to recall a note that the that that our confidence interval for the population mean is given by mu is equal to X plus or minus T subscript alpha divided by 2 multiplied by S divided by the square root of N. So, when we substitute these values back into our equation, we will find that mu is equal to 2.509375 to the power of negative symbol or minus symbol. Plus or minus 3.499 multiplied by 0.01294 divided by the square root of 8. Which will mean that new is going to be equal to 2.509375 to the power of minus or the power of the negative symbol plus or minus 0.01601. Fantastic. So that will mean, and when we make a note, and we'll write our note up above, that our lower bound, which we'll call LB, is going to be equal to 2.4934 g, and our upper bound, which we'll call UB, will be equal to 2.5. 2, so 2.5254 g. Fantastic. So, therefore, as a result, the confidence interval, without a doubt, will be 2.4934 g is less than mu and mu is less than 2.5254 g. And that is our final answer for the interval. Hooray, we did it, and this suggests that the population mean weight hits the specification 2.500 g. So that will mean, when we sum everything together that our final answer, without a doubt, will have to be To quickly recap once again once more. Simply put, our final answer, without a doubt, will have to be. Making sure you can read our answer nice like, oh, there we go. That 2.4934 g is less than me and mu is less than 2.5254 g. And this suggests that the population mean weight hits the specification 2.500 g, and that is our final pair of answers. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Minting Quarters Listed below are weights (grams) of quarters minted after 1964 (based on Data Set 40 “Coin Weights” in Appendix B).

b. Specifications require that the quarters have a weight of 5.670 g. What does the confidence interval suggest about that specification?

Key Concepts

Confidence Interval

Hypothesis Testing

Sample Mean and Standard Deviation

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