1) Here are the IQ test scores of 31 seventh-grade girls in a Midwest school dis

Question

1) Here are the IQ test scores of 31 seventh-grade girls in a Midwest school district: 114 100 104 89 102 91 114 114 103 105 108 130 120 132 111 128 118 119 86 72 111 103 74 112 107 103 98 96 112 112 93 The IQ test scores of 47 seventh-grade boys in the same district are 111 107 100 107 115 111 97 112 104 106 113 109 113 128 128 118 113 124 127 136 106 123 124 126 116 127 119 97 102 110 120 103 115 93 123 79 119 110 110 107 105 105 110 77 90 114 106 (a) Explain how you would check whether using the t-procedures would work here. (no need to actually check (b) Treat these data as SRSs from all seventh-grade students in the district. Is there good evidence that girls and boys differ in their mean IQ scores? State hypotheses, carry out a test, and report your conclusions. (c) Describe what the p-value represents. (d) Use the data in the previous exercise to give a 95% confidence interval for the difference between the mean IQ scores of all boys and girls in the district.

Explanation / Answer

PART A.
data is drawn as a sample,so we conisder this as to be t test

PART B.
Given that,
sample data set
(
114,100,104,89,102,91,114,114,103,105
108,130,120,132,111,128,118,119,86,72
111,103,74,112,107,103,98,96,112,112,93
)
mean(x)=105.84
standard deviation , s.d1=14.27
number(n1)=31

sample data set
(
111,107,100,107,115,111,97,112,104,106,113
109,113,128,128,118,113,124,127,136,106,123
124,126,116,127,119,97,102,110,120,103,115
93,123,79,119,110,110,107,105,105,110,77
90,114,106
)

y(mean)=110.96
standard deviation, s.d2 =12.12
number(n2)=47
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.042
since our test is two-tailed
reject Ho, if to < -2.042 OR if to > 2.042
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =105.84-110.96/sqrt((203.6329/31)+(146.8944/47))
to =-1.6444
| to | =1.6444
critical value
the value of |t | with min (n1-1, n2-1) i.e 30 d.f is 2.042
we got |to| = 1.64442 & | 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 != -1.6444 ) = 0.111
hence value of p0.05 < 0.111,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -1.6444
critical value: -2.042 , 2.042
decision: do not reject Ho
we dont have evidence to support the claim

PART C.
p-value: 0.111

PART D.
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 = [( 105.84-110.96) ± t a/2 * sqrt((203.633/31)+(146.894/47)]
= [ (-5.12) ± t a/2 * 3.114]
= [-11.478 , 1.238]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [-11.478 , 1.238] 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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