Answer:
Step-by-step explanation:
The question is incomplete. The complete question is
Coaching companies claim that their courses can raise the SAT scores of high school students. But students who retake the SAT without paying for coaching also usually raise their scores. A random sample of students who took the SAT twice found 427 who were coached and 2733 who were uncoached. Starting with their verbal scores on the first and second tries, we have these summary statistics:
Try 1 Try 2 Gain
n x s x s x s
Coached 427 500 92 529 97 29 59
Uncoached 2733 506 101 527 101 21 52
Use Table C to estimate a 90% confidence interval for the mean gain of all students who are coached.
at 90% confidence.
Now test the hypothesis that the score gain for coached students is greater than the score gain for uncoached students. Let μ1 be the score gain for all coached students. Let μ2 be the score gain for uncoached students.
(a) Give the alternative hypothesis:
μ1 - μ2.
(b) Give the t test statistic:
(c) Give the appropriate critical value for \alpha =5%:
Solution:
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)
Where
x1 = sample mean score gain for all coached students
x2 = sample mean score gain for all uncoached students
s1 = sample standard deviation score gain of coached students
s2 = sample standard deviation score gain of uncoached students
For a 90% confidence interval, the z score is 1.645
From the information given,
x1 = 29
s1 = 59
n1 = 427
x2 = 21
s2 = 52
n2 = 2733
x1 - x2 = 29 - 21 = 8
z√(s1²/n1 + s2²/n2) = 1.645√(59²/427 + 52²/2733) = 4.97
90% Confidence interval = 8 ± 4.97
a) The population standard deviations are not known. it is a two-tailed test. The random variable is μ1 - μ2 = difference in the score gain for coached and uncoached students.
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
This is a right tailed test.
b) 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)
t = (29 - 21)/√(59²/427 + 52²/2733)
t = 2.65
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 = [59²/427 + 52²/2733]²/[(1/427 - 1)(59²/427)² + (1/2733 - 1)(52²/2733)²] = 9.14/0.1564
df = 58
c) from the t distribution table, the critical value is 1.67
In order to reject the null hypothesis, the test statistic must be smaller than - 1.67 or greater than 1.67
Since - 2.65 < - 1.67 and 2.65 < 1.67, we would reject the null hypothesis.
Therefore, at 5% significance level, we can conclude that the score gain for coached students is greater than the score gain for uncoached students.
