(1 pt) Ten randomly selected people took an IQ test A, and next day they took a
ID: 3319398 • Letter: #
Question
(1 pt) Ten randomly selected people took an IQ test A, and next day they took a very similar IQ test B. Their scores are shown in the table below.
1. Consider (Test A - Test B). Use a 0.05 significance level to test the claim that people do better on the second test than they do on the first.
(a) What test method should be used?
A. Two Sample t test for means
B. Two Sample z test for proportions
C. Matched pairs test
(b) The test statistic is
(c) The critical value is
(d) Is there sufficient evidence to support the claim that people do better on the second test?
A. Yes
B. No
2. Construct a 95% confidence interval for the mean of the differences. Again, use (Test A - Test B).
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Explanation / Answer
1.a.
A. Two Sample t test for means
Given that,
mean(x)=99.9
standard deviation , s.d1=16.5358
number(n1)=10
y(mean)=101.1
standard deviation, s.d2 =16.9145
number(n2)=10
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, = 0.05
from standard normal table,right tailed t /2 =1.833
since our test is right-tailed
reject Ho, if to > 1.833
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =99.9-101.1/sqrt((273.43268/10)+(286.10031/10))
to =-0.16
| to | =0.16
critical value
the value of |t | with min (n1-1, n2-1) i.e 9 d.f is 1.833
we got |to| = 0.16042 & | t | = 1.833
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value:right tail - Ha : ( p > -0.1604 ) = 0.56195
hence value of p0.05 < 0.56195,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 > u2
b.
test statistic: -0.16
c.
critical value: 1.833
decision: do not reject Ho
p-value: 0.56195
d.
No,
there is sufficient evidence to support the claim that people do better on the second test
2.
TRADITIONAL METHOD
given that,
mean(x)=99.9
standard deviation , s.d1=16.5358
number(n1)=10
y(mean)=101.1
standard deviation, s.d2 =16.9145
number(n2)=10
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((273.433/10)+(286.1/10))
= 7.48
II.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, =
from standard normal table, two tailedand
value of |t | with min (n1-1, n2-1) i.e 9 d.f is 2.262
margin of error = 2.262 * 7.48
= 16.92
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (99.9-101.1) ± 16.92 ]
= [-18.12 , 15.72]
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DIRECT METHOD
given that,
mean(x)=99.9
standard deviation , s.d1=16.5358
sample size, n1=10
y(mean)=101.1
standard deviation, s.d2 =16.9145
sample size,n2 =10
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 = [( 99.9-101.1) ± t a/2 * sqrt((273.433/10)+(286.1/10)]
= [ (-1.2) ± t a/2 * 7.48]
= [-18.12 , 15.72]
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interpretations:
1. we are 95% sure that the interval [-18.12 , 15.72] 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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