[DATA] Concrete As concrete cures, it gains strength. The foll...

Question

[DATA] Concrete As concrete cures, it gains strength. The following data represent the 7-day and 28-day strength (in pounds per square inch) of a certain type of concrete:

Table showing paired 7-day and 28-day concrete strength values used for linear regression analysis.

a. Treating the 7-day strength as the explanatory variable, x, determine the estimates of β₀ and β₁.

Accepted Answer

Welcome back everyone. In this problem, a researcher records the following data on the number of orders of student studies per week, X, and their corresponding test scores Y for 5 students. 467, 672, 878, 1085, and 1290. Treating orders studied as the explanatory variable estimate beta knot and beta 1 for the regression line. A says beta knot is 60.5, while beta 1 is 2.5. B says they are 62 and 2 respectively. C 54.8 and 3 respectively, and D 55 and 3.9 respectively. Now, in order for us to estimate beta not and beta one, let's first ask ourselves, what do we know about how they are related to the regression line. Well, we are estimating the regression coefficients, that is beta not the intercept, and beta one, the slope in the equation for the regression line. Recall that. For the regression line, its equation will be ye equals beta0 plus beta 1 X. OK, where the slope beta 1 is equal to the sum of the differences between the X values and their means multiplied by the sum of the differences between the Y values and their means. All divided by the sum of the squares of the differences between the X values and their means. Next, we also know that beta not or Y intercept will be equal to the mean of Y minus beta 1 multiplied by the mean of X. So in this case, if we can figure out the meaning of the X and Y values, the differences between the the X and Y values and their means, and the square of the difference between the X values and their mean. Then we should be able to plug it into our formulas and thus solve for beta one and beta not. So first, let's start by calculating the means, OK? So now calculating the means. The mean value of X will be equal to 1 divided by n, the sample size, multiplied by the sum of the X values. OK. No, when we calculate that, OK, it's equal to 4 + 6 + 8 + 10 + 12. All our x values divided by 5, which equals 8. On the other hand, for the mean for the Y values, that's one divided by and multiplied by the sum of y values. Which will be equal to 67 + 72 + 78 + 85 plus 90 divided by 5. And that equals 78.4. Know that we have our respective means, we can calculate the rest of values we need. So I think to make this easiest, let's represent this in a table. So we have our X values, our y values, the differences for the X values and their means, the differences for the Y values and their means, and then we have the differences for the X values and their means squared. And what we're gonna do is that if you recall from our equation for beta one, we need to find the sum of the X differences, the Y differences, and the sum of the squares for the X differences. So we'll add them up once we've calculated. So now, for X and Y values, we know based on our problem, we have 4 and 67 as one pair, uh 6 and 72 as the other. 878 as the 3rd, 1085 as the 4th, and 12 and 90 as the 5th. So, starting with our values for uh the difference between X and their means. In the first one, for example, that is going to be 4 minus 8, OK, which is going to be equal to 4, or -4, rather. Uh, for 6, sorry, for 6, that's gonna be equal to -2, 6 minus 8. For 8, that's going to be 0, for 10, that's going to be 2, and for 12, that's going to be 4, OK. Now for our Y values, next, we can calculate those differences as well. Which when Y equals 67 is going to be uh negative 11.4, OK. When Y is 72, that will be negative 6.4, because remember we're subtracting 78.4. When Y is 78, that will be negative 0.4. Uh, when it's 10, sorry, when it's 85, that will be 6.6, and when it's 90, that's 11.6, OK? So you know what, let me just make some space here because we should we we're finding the products of the difference for X values and the difference for the Y values, OK? So now when we find a product, let me just write that here. So that's we're multiplying these two. So that's XI minus X bar. Multiplied by YI minus YR OK. Then, for our first term, -4 multiplied by -11.4, that's 45.6. Next, we get 12.8, 0, OK, 13.2, and 46.4. And know if we square our differences for the X terms and the means, then we'll end up with 16 for the first term, then 404, and then 16. So now let's find our sons, OK? Now we're gonna find our sums. How can we get our sons? Well, to find the sums of our products, when we add all of these up, we get it to be equal to 118.0. And for differences for squares, the sum will be 40. So now that we have these sums, we can plug them into the slope, beta one, OK, to get the calculated values. Because no, this means that beta one is going to be equal to. Or some here or or products which is 118 divided by the sum of the squares which is 40, which will be equal to 2.95 or approximately 3 to 1 decimal place. I know that we have that, this means that our intercept beta not is going to be equal to the the mean for Y 78.4, OK, minus beta 1, which is 2.95, multiplied by or X mean, which is 8. And when we calculate here, we get that to be 54. Point it Therefore, this tells us that our slope is 3 and our intercept is 54.8. If we look back at our answer choices, that means C is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

[DATA] Concrete As concrete cures, it gains strength. The following data represent the 7-day and 28-day strength (in pounds per square inch) of a certain type of concrete:

a. Treating the 7-day strength as the explanatory variable, x, determine the estimates of β₀ and β₁.

Key Concepts

Simple Linear Regression

Least Squares Estimation

Explanatory and Response Variables

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