Using the information provided in problem one, develop a 95% confidence interval
ID: 3322385 • Letter: U
Question
Using the information provided in problem one, develop a 95% confidence interval for the difference in the mean SAT math scores between Westridge and Eastridge students. Interpret your confidence interval (i.e. what does it mean in this particular case?). Round to 4 (FOUR) decimal places.
****Information from problem 1--à“A school board member for Westridge, a large suburban school district, suspects his district's high school students are performing lower on the math section of the SAT test than those of crosstown rival, Eastridge. To test his belief, he randomly samples the SAT math section scores of 45 current students from Westridge and 43 current students from Eastridge. For Westridge, the sample mean score was 492 with a sample standard deviation of 37. For Eastridge, the sample means core was 511 with a sample standard deviation of 41. At the .05 level of significance, is there enough evidence to conclude that the mean score on the SAT score in the math section is lower for Westridge students than it is for Eastridge students? Assume SAT scores are normally distributed. List and clearly label all eight steps. Round to 4 (FOUR) decimal places.”
Explanation / Answer
PART A.
Given that,
mean(x)=492
standard deviation , s.d1=37
number(n1)=45
y(mean)=511
standard deviation, s.d2 =41
number(n2)=43
null, Ho: u1 = u2
alternate, H1: u1 < u2
level of significance, = 0.05
from standard normal table,left tailed t /2 =1.682
since our test is left-tailed
reject Ho, if to < -1.682
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =492-511/sqrt((1369/45)+(1681/43))
to =-2.2788
| to | =2.2788
critical value
the value of |t | with min (n1-1, n2-1) i.e 42 d.f is 1.682
we got |to| = 2.27884 & | t | = 1.682
make decision
hence value of | to | > | t | and here we reject Ho
p-value:left tail - Ha : ( p < -2.2788 ) = 0.01391
hence value of p0.05 > 0.01391,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 < u2
test statistic: -2.2788
critical value: -1.682
decision: reject Ho
p-value: 0.01391
there is enough evidence to conclude that the mean score on the SAT score in the math section
is lower for Westridge students than it is for Eastridge students
PART B.
TRADITIONAL METHOD
given that,
mean(x)=492
standard deviation , s.d1=37
number(n1)=45
y(mean)=511
standard deviation, s.d2 =41
number(n2)=43
I.
stanadard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
stanadard error = sqrt((1369/45)+(1681/43))
= 8.338
II.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, = 0.05
from standard normal table, two tailed and
value of |t | with min (n1-1, n2-1) i.e 42 d.f is 2.018
margin of error = 2.018 * 8.338
= 16.825
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (492-511) ± 16.825 ]
= [-35.825 , -2.175]
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DIRECT METHOD
given that,
mean(x)=492
standard deviation , s.d1=37
sample size, n1=45
y(mean)=511
standard deviation, s.d2 =41
sample size,n2 =43
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 492-511) ± t a/2 * sqrt((1369/45)+(1681/43)]
= [ (-19) ± t a/2 * 8.338]
= [-35.825 , -2.175]
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interpretations:
1. we are 95% sure that the interval [-35.825 , -2.175] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population proportion
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