You have two categories but three columns?
Find the means first = sum of scores/number of scores
Subtract each of the scores from its mean and square each difference. Find the sum of these squares. Divide that by the number of scores to get variance.
Standard deviation = square root of variance
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√n
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z score.
90% = mean ± 1.645 SD (This is the Z score for .45 from the mean found in that same table.
Composite ACT Scores
Non-Athletes Athletes
25 21 22
22 27 21
19 29 24
25 26 27
24 30 19
25 27 23
24 26 17
23 23
a. Test an appropriate hypothesis and state your conclusion. Use alpha = .05 for your test. Show the null and alternative hypothesis, the p-value, your conclusion to reject or not, and finally a summarizing statement regarding your conclusion.
Then create a 90% confidence interval and tell me what it means in terms of the problem.
2 answers
SO FAR I HAVE 9.63 FOR THE 1ST 2 COLUMNS
AND 7.56 FOR LAST COLUMN FOR THE VARIANCES. I GOT Z= .0875
SEDIFF =1.2364 SEm=.6019 for 1st set of #s and SEm=1.08 for 2nd set of #s
is that correct and now what?
AND 7.56 FOR LAST COLUMN FOR THE VARIANCES. I GOT Z= .0875
SEDIFF =1.2364 SEm=.6019 for 1st set of #s and SEm=1.08 for 2nd set of #s
is that correct and now what?