The scores represent IQ scores for 31 seventhgrade girls and

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

Data is at the bottom

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 should use JMP to find any summary statistics such as the mean, standard deviation, etc., you should perform the test by hand and verify the result using JMP/other stat software).

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

IQ Scores

Gender	Iqscores
Female	114
Female	100
Female	104
Female	89
Female	102
Female	91
Female	114
Female	114
Female	103
Female	105
Female	108
Female	130
Female	120
Female	132
Female	111
Female	128
Female	118
Female	119
Female	86
Female	72
Female	111
Female	103
Female	74
Female	112
Female	107
Female	103
Female	98
Female	96
Female	112
Female	112
Female	93
Male	111
Male	107
Male	100
Male	107
Male	115
Male	111
Male	97
Male	112
Male	104
Male	106
Male	113
Male	109
Male	113
Male	128
Male	128
Male	118
Male	113
Male	124
Male	127
Male	136
Male	106
Male	123
Male	124
Male	126
Male	116
Male	127
Male	119
Male	97
Male	102
Male	110
Male	120
Male	103
Male	115
Male	93
Male	123
Male	79
Male	119
Male	110
Male	110
Male	107
Male	105
Male	105
Male	110
Male	77
Male	90
Male	114
Male	106

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]

The scores represent IQ scores for 31 seventh-grade girls and 47 seventh-grade boys in the same Midwestern school district. Data is at the bottom a) Construct n

The scores represent IQ scores for 31 seventhgrade girls and

Solution

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