The scores represent IQ scores for 31 seventh-grade girls and 47 seventh-grade b

Question

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

Explanation / Answer

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]

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