The following data were collected from a repeated measures...

The following data were collected from a repeated-measures study. Perform an ANOVA. Use a .05 alpha level. (10 POINTS) Treatment A B _____________ G = 36 2 6 2 6 ?X2 = 192 4 4 4 8 ______________ T = 12 T = 24 SS = 4 SS = 8 1) State the null hypothesis. 2) Calculate necessary statistics (USE AN ANOVA even though K=2, and ??= .05). 3) Make a decision about your null hypothesis and explain what that decision means. 4) Make a conclusion about your IV and DV that includes an APA format summary of your statistical analysis.

Transcript

00:01 Okay, so here let the null hypothesis would be h0 and this is mu1 will be equal to m times mu 2 okay and there is no difference between the means of the two treatments a and b okay and the alternative hypothesis let's write it here so it will be clear and this is m times mu 2 okay and this is equal to mu1 okay and the alternative hypothesis h sub a will be me one will be not equal to m mu two okay and there is no difference between the mean of the two treatments a and b okay now in part two we are given that k is equal to two and alpha is a point zero five okay and the the p value is calculated for p is equal to nine this p value is so so per degree of freedom is equal to one and degree of freedom one uh two is six okay so par critical value it is calculated that alpha is uh so we are degree of freedom is uh so let let write it here right so let clear it uh let over mean is our mean is and this is equal to 12 plus 24 divided by 8 okay and this is 36 divided by 8 so this is 4 .5 and the degrees of freedom 1 is k minus 1 so this will be 2 minus 1 or this is 1 okay and the ss treatment is let the ss treatment will be the s treatment is equal to the sum up n times x bar up i minus overall overall mean square okay so this is equal to four times three minus 4 .5 whole square plus four times six minus four point five whole square okay so this is equal to four times minus one point five square plus four times one point five square okay so this is four times two point two five okay so nine plus nine plus nine plus nine plus nine will be equal to 18 right and m s t r is equal to s and this will be equal to treatment divided by degrees of freedom 1 so this is 18 divided by 1 this is 18 right and the s s error is equal to sum times the sum of squares okay sum of squares so this is a point plus 8 this is 12 right and the degrees of freedom 2 this is equal to n minus k right and n is 8 minus 2 is 6 so the ms error is this is equal to the ss error okay divided by degrees of freedom 2 now this will be equal to 12 divided by 6 so this is 2 right and if is equal to m s t r divided by m s e right so this will be equal to 18 divided by 2 this is 9 right and so if is equal to 9 now let write it here right so this is the source this is an o right so this is between and this is within so this will be more easy right and this is total right so this is s and this is degrees of freedom and this is mean square mean square and if and if cv and p right so this is 18 this is 12 and the total is 30 30 and d is 167 right and the mean square is 18 2 and if is 9 and fcv is 5 .9 okay and p is 0 .0 .040 okay and p is 0 .040 so so in three, so the decision is, so since a test of a is, let's write it here, so if test of a is, this is greater than a critical, right than a critical of 5 .98...

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