Consider the data of Review Exercise 13.58 on page 568 . Make significance tests on the followin

Consider the data of Review Exercise 13.58 on page 568 ....

Key Concepts

Analysis of Variance (ANOVA)

ANOVA is a statistical method used to test for differences among three or more group means by partitioning the total variability into components attributable to the treatments and to random error. It provides the framework within which contrasts are evaluated to understand specific differences between groups.

Contrasts in ANOVA

Contrasts are specific linear combinations of group means, with coefficients that sum to zero, designed to test focused hypotheses about differences among groups. They allow researchers to make targeted comparisons rather than relying on overall omnibus tests.

Contrast Coefficients

Contrast coefficients are the weights assigned to each group mean within a contrast. They are selected so that their sum is zero, which ensures that the contrast focuses solely on the differences among groups rather than on the individual group means.

Significance Testing for Contrasts

This concept involves testing the null hypothesis that a specified linear combination of group means (the contrast) is zero. The test is built on the estimated contrast and its standard error, which is calculated using the within-group variability from the ANOVA, and it typically involves a t-test or F-test.

t-test for Contrasts

A t-test for contrasts computes a t-statistic by dividing the estimated contrast by its standard error. This procedure allows for determining whether the specific combination of group means is significantly different from zero, thereby assessing the significance of the hypothesized differences.

Transcript

00:02 Okay, the question says, consider the data of review exercise dating .58 on page 568, make significance test on the following contrast.

00:14 A, b versus a, c and d, and then b, c versus a and d.

00:23 And then we have c, which is a versus d.

00:28 So let's go to exercise 13 .58 and c, the value.

00:33 18 .58 right 13 .58 it says four laboratories are being used to perform chemical analysis samples of the same material are sent to the laboratories for analysis as part of a study to determine whether or not they give or on the average the same results.

00:53 The analytical results for the four laboratories are shown so with a, b, c and d these are those of laboratories and then the results are given there.

01:03 They are five per each group there so we have 20 datasets four times five yes we have 20 data sets there then maybe to read one of the questions that it was asked there though we are not answering this one we are supposed to do our own comparisons there so it says use partless tests to show that they would that the within laboratory variances are not significantly different at the 0 .05 level of significance all right now let's go to our own right so i've just done the presentation here so i need to explain to you how we literally get these values okay right and then eventually we have to explain again how we got the f the computed values and to compare and we do the comparisons there right so maybe let's start by seeing b versus c versus a c and p to find the sum of squares so to find the sum of squares let me go down into the working there i'll come and show which you see to qualify the so to find the sum of squares of say b versus ac and d there so what we do is we have to find the sum of squares of si right the sum of squares of si which is right n then si over the summation of the coefficient but they have to be squared so meaning if we are to use this formula first of all we need to find the size squared okay so we need to find si so let's find size sign we find si sign is in psi sign we find by saying the summation of the coefficients right multiplied by the mean of each group there so the mean of each group so for b let me say let me put first of all put the coefficients for b right a c and d the coefficients i gave this one three and this one minus one minus one and minus one so that they complement each other they might if you add them they must get a zero must get a zero there so now let's see let's calculate the side let me quote down here we have to calculate the side for that okay the side for that so it will be three it's the summation three times the mean of p the mean of the mean of uh column b was a 61 .96 right minus then one times uh 59 that is column a 59 comma 66 then minus one times column c the mean is 6 1 comma 2 2 then minus 1 times if we column d which is okay i'm advice this is supposed to be d this is supposed to be d d nonetheless we just exchange it doesn't affect anything so it's going to be 56 .52 then 56 .5 2 so if you do this computation you get 8 .58 this is our side so we'll need to to then come back here into the actual equation actually the actual calculation that we're making is this one so our end the number of data sets within a group there there are five so it's five multiplied by size squared which is eight comma five eight squared they divided by right this is the summation of the coefficients but squared so it's going to be three squared plus minus one squared plus minus one squared plus minus one squared okay so this is the same as saying if you do this computation what do you get there with a calculator there you get 30 comma 6 735 okay so that's the value that we get for our sum of squares of the first one there let's check if we got it right yes 30 comma 6 7 35 there all right so we are we are okay right let's check again for the calculation of the second one the second one is what um the second one is a c versus c versus a d d is d not r so our side let's couple of the value of si, si would be, right.

06:24 Now that you've put the formulas up there, we're just going to put the values for si.

06:29 Right.

06:31 Maybe let me, first of all, give you the coefficients.

06:35 It's c, a, and d.

06:41 Right.

06:42 Our c is two, negative one, negative one.

06:45 Okay, so our our size our size is going to be two brackets 56 .5 to the 56 .5 to 5 to 5 minus okay i'm not going to put minus 1 because it becomes minus 161 12 minus 59 comma 66 and what we get with the calculator there get minus 7 .74 right now let's calculate let's find our sum of si right sum of size is going to be five okay i'm writing the formula i wanted to five minus seven comma seven squared all over this is two squared plus one plus one so it's four plus six so it's over six there right this is a squared there so if you do that what do we get we get four 49 .923 there...

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