1a.) One sample has n = 17 with SS = 1646, and a second sample has n = 17 with S

Question

1a.) One sample has n = 17 with SS = 1646, and a second sample has n = 17 with SS = 1710. (a) Find the pooled variance for the two samples. (Use/round to 2 decimal places.) Pooled variance = ______________

1b.) Compute the estimated standard error for the sample mean difference. (Use/round to 3 decimal places.) Estimated standard error = ___________

1c.) If the sample mean difference is 6 points, is this enough to reject the null hypothesis and conclude that there is a significant difference for a two-tailed test at the 0.10 level? (Use/round to 3 decimal places.)

t-critical = ± __________ t = __________

1d.) What is the decision regarding the null? Accept the null or reject it? Write "accept" or "reject" in the blank. __________

1e.) Calculate the percentage of variance accounted for (r2) to measure the effect size for an 6-point mean difference (Use/round to 3 decimal places). r2 = _____________

1f.) What size is the effect? Write "small", "medium", or "large" in the blank. __________

Explanation / Answer

(1a)pooled variance=(SS1+SS2)/(n1+n2-2)=(1646+1710)/(17+17-2)=104.875

(1b)pooled standard deviation=sp=sqrt(104.875)=10.2408

standard error for the sample mean difference=((sp*(1/n1 +1/n2)1/2) =10.2408*sqrt(1/17 + 1/17)=3.5126

(1c)) t=(mean1-mean2)/((sp*(1/n1 +1/n2)1/2) =sample mean difference/SE(difference)=6/3.5126=1.708

with (n1+n2-2)=(17+17-2)=32 df

critical t(0.1,32)=1,693

(1d) reject H0, as ciritcal t=1.693 is less than calcuated t=1.708

(1e) here not possible to calcuate r2,

still we can calculate effect size

cohen'd effect size=sample mean difference/pooled standard deviation=6/10.2408=0.586

(1f) this is large effect size

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