Ten randomly selected people took an IQ test A, and next day they took a very si

Question

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. Person A B C D E F G H I J Test A 107 98 88 82 102 71 124 117 112 95 Test B 113 100 89 83 100 74 126 122 112 93 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. (Note: You may wish to use software.) (a) What test method should be used? A. Two Sample t B. Matched Pairs C. Two Sample z (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). <

Explanation / Answer

a.
t test for difference of mean unequal
Given that,
mean(x)=99.6
standard deviation , s.d1=16.3245
number(n1)=10
y(mean)=101.2
standard deviation, s.d2 =16.9758
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 =99.6-101.2/sqrt((266.4893/10)+(288.17779/10))
to =-0.215
| to | =0.215
critical value
the value of |t | with min (n1-1, n2-1) i.e 9 d.f is 1.833
we got |to| = 0.21483 & | t | = 1.833
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value:right tail - Ha : ( p > -0.2148 ) = 0.58266
hence value of p0.05 < 0.58266,here we do not reject Ho
ANSWERS
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b.
null, Ho: u1 = u2
alternate, H1: u1 > u2
c.
test statistic: -0.215
critical value: 1.833
decision: do not reject Ho
p-value: 0.58266
d.
no,
we do not have enough evidence to support the claim that people do better on the second test

e.
TRADITIONAL METHOD
given that,
mean(x)=99.6
standard deviation , s.d1=16.3245
number(n1)=10
y(mean)=101.2
standard deviation, s.d2 =16.9758
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((266.489/10)+(288.178/10))
= 7.448
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 * 7.448
= 16.846
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (99.6-101.2) ± 16.846 ]
= [-18.446 , 15.246]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=99.6
standard deviation , s.d1=16.3245
sample size, n1=10
y(mean)=101.2
standard deviation, s.d2 =16.9758
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 = [( 99.6-101.2) ± t a/2 * sqrt((266.489/10)+(288.178/10)]
= [ (-1.6) ± t a/2 * 7.448]
= [-18.446 , 15.246]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [-18.446 , 15.246] 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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