The scores represent IQ scores for 31seventhgrade girls and

The scores represent IQ scores for 31seventh-grade girls and 47 seventh-grade boys in the same Midwestern school district.

a. Construct normal probability plots for each group.

b. For each group, use technology to find the mean and standard deviation for each group.

c. Assuming the assumptions are satisfied,what is the name of the test you would use for this problem? (Your options are one sample z-test for the mean, one sample t-test for the mean, one sample z-test for the proportion, matched-pairs t-test for the mean, two-sample (independent) t-test for the mean.)

d. Using ? = 0.05, conduct the hypothesis test used to determine whether there is good evidence that girls and boys differ in the their mean IQ scores. (Although you can use R/Excel to find any summary statistics such as the mean, standard deviation, etc., you should perform the test by hand and verify the result using R/Excel, i.e. write down the equations for the statistics you used in the test).

e. Construct a 95% confidence interval for the difference between the mean IQ scores of all boys and all girls in the district.

Solution

Set Up Hypothesis
Null Hypothesis , There Is No-Significance between them Ho: u1 = u2
Alternate Hypothesis, There Is Significance between them - H1: u1 != u2
Test Statistic
X(Mean)=105.834
Standard Deviation(s.d1)=14.272 ; Number(n1)=31
Y(Mean)=110.956
Standard Deviation(s.d2)=12.1207; Number(n2)=47
we use Test Statistic (t) = (X-Y)/Sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =105.834-110.956/Sqrt((203.68998/31)+(146.91137/47))
to =-1.645
| to | =1.645
Critical Value
The Value of |t | with Min (n1-1, n2-1) i.e 30 d.f is 2.042
We got |to| = 1.64488 & | t | = 2.042
Make Decision
Hence Value of |to | < | t | and Here we Do not Reject Ho
P-Value: Two Tailed ( double the one tail ) - Ha : ( P != -1.6449 ) = 0.11
Hence Value of P0.05 < 0.11,Here We Do not Reject Ho

[ANSWERS]
b)
FEMALE GROUP
X(Mean)=105.834
Standard Deviation(s.d1)=14.272 ; Number(n1)=31

MALE GROUP
Y(Mean)=110.956
Standard Deviation(s.d2)=12.1207; Number(n2)=47

c)
two-sample (independent) t-test for the mean.)

d)
We don\'t have evidence to determine there is good evidence that girls
and boys differ in the their mean IQ scores

e)
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
Where,
x1 = Mean of Sample 1, x2 = Mean of sample2
sd1 = SD of Sample 1, sd2 = SD of sample2
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x1)=105.834
Standard deviation( sd1 )=14.272
Sample Size(n1)=31
Mean(x2)=110.956
Standard deviation( sd2 )=12.1207
Sample Size(n1)=47
CI = [ ( 105.834-110.956) ±t a/2 * Sqrt( 203.689984/31+146.91136849/47)]
= [ (-5.12) ± t a/2 * Sqrt( 9.7) ]
= [ (-5.12) ± 2.042 * Sqrt( 9.7) ]
= [-11.48 , 1.24]