Answer:
Step-by-step explanation:
Part A
For Athletic scholarship,
Mean = (16 + 24 + 20 + 25 + 24 + 23 + 21 + 22 + 20 + 20)/10 = 21.5
Standard deviation = √(summation(x - mean)²/n
n = 10
Summation(x - mean)² = (16 - 21.5)^2 + (24 - 21.5)^2 + (20 - 21.5)^2 + (25 - 21.5)^2 + (24 - 21.5)^2 + (23 - 21.5)^2 + (21 - 21.5)^2 + (22 - 21.5)^2 + (20 - 21.5)^2 + (20 - 21.5)^2 = 64.5
Standard deviation = √64.5/10 = 2.54
For non athletic scholarship,
Mean = (23 + 25 + 26 + 30 + 32 + 26 + 28 + 29 + 26 + 27 + 29 + 27 + 22 + 24 + 25)/15 = 26.6
n = 15
Summation(x - mean)² = (23 - 26.6)^2 + (25 - 26.6)^2 + (26 - 26.6)^2 + (30 - 26.6)^2 + (32 - 26.6)^2 + (26 - 26.6)^2 + (28 - 26.6)^2 + (29 - 26.6)^2 + (26 - 26.6)^2 + (27 - 26.6)^2 + (29 - 26.6)^2 + (27 - 26.6)^2 + (22 - 26.6)^2 + (24 - 26.6)^2 + (25 - 26.6)^2 = 101.6
Standard deviation = √101.6/15 = 2.6
This is a test of 2 independent groups. The population standard deviations are not known. it is a two-tailed test. Let 1 be the subscript for scores of athletes and 2 be the subscript for scores of non athletes.
Therefore, the population means would be μ1 and μ2
The random variable is x1 - x2 = difference in the sample mean scores of athletes and non athletes.
We would set up the hypothesis.
The null hypothesis is
H0 : μ1 = μ2 H0 : μ1 - μ2 = 0
The alternative hypothesis is
H1 : μ1 ≠ μ2 H1 : μ1 - μ2 ≠ 0
Since sample standard deviation is known, we would determine the test statistic by using the t test. The formula is
(x1 - x2)/√(s1²/n1 + s2²/n2)
From the information given,
x1 = 21.5
x2 = 26.6
s1 = 2.54
s2 = 2.6
n1 = 10
n2 = 15
t = (21.5 - 26.6)/√(2.54²/10 + 2.6²/15)
t = - 4.65
The formula for determining the degree of freedom is
df = [s1²/n1 + s2²/n2]²/(1/n1 - 1)(s1²/n1)² + (1/n2 - 1)(s2²/n2)²
df = [2.54²/10 + 2.6²/15]²/[(1/10 - 1)(2.54²/10)² + (1/15 - 1)(2.6²/15)²] = 1.2008/0.1039
df = 12
We would determine the probability value from the t test calculator. It becomes
p value = 0.00056
Since alpha, 0.1 > than the p value, 0.00056, then we would reject the null hypothesis.
Therefore, these data provide convincing evidence of a difference in ACT scores between athletes and nonathletes.
Part B
The formula for determining the confidence interval for the difference of two population means is expressed as
Confidence interval = (x1 - x2) ± z√(s²/n1 + s2²/n2)
For a 90% confidence level, the z score from the normal distribution table is 1.645
x1 - x2 = 21.5 - 26.6 = - 5.1
√(s1²/n1 + s2²/n2) = √(2.54²/10 + 2.6²/15) = 1.05
The confidence interval is - 5.1 ± 1.05
This analysis provides evidence that the mean scores for non athletes is higher than the mean scores for athletes, and that the difference between means in the population is likely to be between 4.05 and 6.15
, you have the remainder term approaching 
, so the horizontal asymptote is the line 
.
