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.

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.

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?

To test an appropriate hypothesis and state the conclusion, we can use the t-test for independent samples.

Null hypothesis (H0): There is no significant difference in composite ACT scores between non-athletes and athletes.
Alternative hypothesis (Ha): There is a significant difference in composite ACT scores between non-athletes and athletes.

Significance level (alpha) = 0.05

To perform the t-test, we need to calculate the mean, standard deviation, and sample size for each group.

Non-Athletes:
Mean (x̄1) = (25 + 22 + 19 + 25 + 24 + 25 + 24 + 23) / 8 = 23.75
Standard Deviation (s1) = √[((25-23.75)^2 + (22-23.75)^2 + (19-23.75)^2 + (25-23.75)^2 + (24-23.75)^2 + (25-23.75)^2 + (24-23.75)^2 + (23-23.75)^2) / (8-1)] ≈ 2.12
Sample Size (n1) = 8

Athletes:
Mean (x̄2) = (21 + 27 + 29 + 26 + 30 + 27 + 26 + 23) / 8 = 26
Standard Deviation (s2) = √[((21-26)^2 + (27-26)^2 + (29-26)^2 + (26-26)^2 + (30-26)^2 + (27-26)^2 + (26-26)^2 + (23-26)^2) / (8-1)] ≈ 2.68
Sample Size (n2) = 7

Next, we can calculate the t-test statistic and the p-value using a statistical software or t-tables. Assuming the sample sizes are small (<30) and the distributions are not known to be normal, we can use the Welch's t-test, which accounts for unequal variances.

t-test statistic = (x̄1 - x̄2) / √(s1^2 / n1 + s2^2 / n2)
p-value = p(t > t-test statistic)

Once we have the p-value, we compare it to the significance level (alpha) to assess the statistical significance.

If the p-value is less than alpha (i.e., p-value < alpha), we reject the null hypothesis and conclude that there is a significant difference in composite ACT scores between non-athletes and athletes.

Finally, to create a 90% confidence interval, we can use the following formula:

Confidence interval = (x̄1 - x̄2) ± t *
√(s1^2 / n1 + s2^2 / n2)

Where t is the critical value from the t-distribution table corresponding to a 90% confidence level and degrees of freedom approximated using a Welch-Satterthwaite equation.

The confidence interval represents a range of values within which we can be 90% confident that the true difference in composite ACT scores between non-athletes and athletes lies. If this interval does not contain zero, it provides evidence of a significant difference between the two groups.