Question

20. An educational psychologist wanted to determine if 7-year old girls perform significantly morent than 7-year old boys on a visual perception task. Five girls and 5 boys were randomly selected from the entire set of 7-year olds in a school. Each were then administered a visual perception test. Higher scores indicate better visual perception. Sum of squares (SS) for girls is 22 and SS for boys is 6.8. Girls 17 16 21 15 | 16 SS = 22 Boys 1 12 12 15 14 13 SS = 6.8 a. Conduct a two-tailed test with a = .05 to determine if the psychologist's hypothesis was correct. (1 pt.) b. Construct a 95% confidence interval for the two sample mean difference. (1 pt.) C. Is the confidence interval result consistent with your hypothesis testing decision? Please explain. (0.5 pt.)

Answer 1

20.
a.
Given that,
mean(x)=17
standard deviation , s.d1=2.3452
number(n1)=5
y(mean)=13.2
standard deviation, s.d2 =1.3038
number(n2)=5
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.776
since our test is two-tailed
reject Ho, if to < -2.776 OR if to > 2.776
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =17-13.2/sqrt((5.49996/5)+(1.69989/5))
to =3.1667
| to | =3.1667
critical value
the value of |t α| with min (n1-1, n2-1) i.e 4 d.f is 2.776
we got |to| = 3.1667 & | t α | = 2.776
make decision
hence value of | to | > | t α| and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 3.1667 ) = 0.034
hence value of p0.05 > 0.034,here we reject Ho
ANSWERS
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null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 3.1667
critical value: -2.776 , 2.776
decision: reject Ho
p-value: 0.034
we have enough evidence to support the claim that difference in means between girls and boys.
b.
TRADITIONAL METHOD
given that,
mean(x)=17
standard deviation , s.d1=2.3452
number(n1)=5
y(mean)=13.2
standard deviation, s.d2 =1.3038
number(n2)=5
I.
standard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
standard error = sqrt((5.5/5)+(1.7/5))
= 1.2
II.
margin of error = t a/2 * (standard error)
where,
t a/2 = t -table value
level of significance, α = 0.05
from standard normal table, two tailed and
value of |t α| with min (n1-1, n2-1) i.e 4 d.f is 2.776
margin of error = 2.776 * 1.2
= 3.331
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (17-13.2) ± 3.331 ]
= [0.469 , 7.131]
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DIRECT METHOD
given that,
mean(x)=17
standard deviation , s.d1=2.3452
sample size, n1=5
y(mean)=13.2
standard deviation, s.d2 =1.3038
sample size,n2 =5
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 = [( 17-13.2) ± t a/2 * sqrt((5.5/5)+(1.7/5)]
= [ (3.8) ± t a/2 * 1.2]
= [0.469 , 7.131]
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interpretations:
1. we are 95% sure that the interval [0.469 , 7.131] 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
c.
yes,
the confidence interval is consistent with the hypothesis test.