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Three students, Linda, Tuan, and Javier, are given five laboratory rats each for a nutritional experiment. Each rat's weight is recorded in grams. Linda feeds her rats Formula A, Tuan feeds his rats Formula B, and Javier feeds his rats Formula C. At the end of a specified time period, each rat is weighed again, and the net gain in grams is recorded. Using a significance level of 10%, test the hypothesis that the three formulas produce the same mean weight gain.$$\begin{array}{|c|c|c|}\hline \text { Linda's rats } & {\text { Tuan's rats }} & {\text { Javier's rats }} \\ \hline 43.5 & {47.0} & {51.2} \\ \hline 39.4 & {40.5} & {40.9} \\ \hline 41.3 & {38.9} & {37.9} \\ \hline 46.0 & {46.3} & {45.0} \\ \hline 38.2 & {44.2} & {48.6} \\ \hline\end{array}$$

a) The null hypothesis that three mean rat weight changes caused by different feed formulas are the same is given as below:$H_{0} : \mu_{1}=\mu_{2}=\mu_{3}$b) The alternate hypothesis will be at least any two of the means are different.c) The degree of freedom in the numerator is $2,$ and the degree of freedom in the denominator is $12 .$d) The $F$ distribution is used for the test of three different means.e) 0.6685f) 0.53g) see graphh) a.Level of significance $\alpha$ is 0.10b. Decision: Null hypothesis will be accepted.c. Reason for decision: P-value is 0.53 which is greater than the 10 level of significance.d. Three formulas produce the same mean weight gain.

Intro Stats / AP Statistics

Chapter 13

F Distribution and One-Way ANOVA

Section 3

Facts About the F Distribution

Distribution and One-Way ANOVA

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we're gonna press enter, um, And then we get the three outputs, right? We get that f is we're gonna have equal to end somewhere numbers, and then we get P is equal to about 0.53 and it keeps going. And then the fact that because they would get the degrees of freedom DF, um, and and so on. So, um, then for part A. Um Okay, well, the null hypothesis that the three, um mean commuting manages are the same is given as well. That's gonna be our null hypothesis. H sub zero is gonna be, um, you won being equal to again. This is not a youth today, Mu. This is Mu Mu One is equal to, um, you two, which is equal to, um you three. Okay. And then our our our our alternative hypothesis for part beat will be Well, at least two of the means are different. So we say that the alternative the alternative hypothesis, uh, will be that at least a T least any two of the means. Oh, crap. At least any two of the means are different. Okay, um then for C Well, um, the degree of freedom in the numerator is too. So the degree of freedom degree of freedom in the numerator mhm is to, um, and the degree of freedom and the degree of freedom in the denominator we see is 12. Um, then we go on to part deep. So indeed, the f distribution here is used. We see again, and I calculated we see that the F distribution is used for the test of the three different means. So the F distribution AB distribution here is used, um, and then for e b test statistic. So the test statistic statistic, um, is given as 0.6 685 and the P value is 0.53 So we have for F three p value. Mm. Um, for the test is I'm still 0.53 Yeah. Okay. And then, um, Fergie while we get the graph off the distribution. So Fergie here, our our graph is gonna look, um, something like this we have. This is our It's your density. This is our ex. Um, So we come out from see this here zero to the 0.6686 which is here, and a graph is gonna look something like this. We then become a satanic here. And, um so it's Well, maybe this actually should be moved a little bit. Well, in the just the 0.6686 maybe we're right about that. Save it about here, actually. Okay. And then way Care about this area here? Um, And, um, for H, the level of significance here is, well, 0.10 So we say that I'm still in red here way. Say that the level of significance Alfa is 0.1. No, um, and our decision here will be that are novel hypothesis null hypothesis. Well, then well will be accepted. So our decision, um, is that the null hypothesis? Cool. Um, will be accepted. Okay. And what's the reason for this decision? Well, the reason would be that the p value is 0.53 which is greater than our, um, level of significance. So the reason for decision our reason is that the p value, um, which is 0.53 is greater than greater than Alfa, right. Greater than Alfa, which is our 0.130 yeah. And we can then say that the 34 The three formulas produce the same mean weight length. So we can say that, um, the three formulas produce the same mean wait.

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