Gender bias in textbooks. To what extent do syntax textbooks, which analyze the

Question

Gender bias in textbooks. To what extent do syntax textbooks, which analyze the structure of sentences, illustrate gender bias? A study of this question sampled sentences from 10 texts One part of the study examined the use of the words "girl "boy, " "man, " and "woman. We will call the first two words juvenile and the last two adult. Is the proportion of female references that are juvenile (girl) equal to the proportion of male references that are juvenile (boy)? Here are data from one of the texts: (a) Find the proportion of juvenile references for females and its standard error. Do the same for the males. (b) Give a 90% confidence interval for the difference and briefly summarize what the data show. (c) Use a test of significance to examine whether the two proportions are equal.

Explanation / Answer

PART A.
Proportion 1
No. of chances( X1 )=48
No.Of Observed (n1)=60
P1= X1/n1=0.8
Proportion 2
No. of chances(X2)=52
No.Of Observed (n2)=132
P2= X2/n2=0.394

PART B.
Margin of Error = Z a/2 Sqrt(p1(1-p1)/n1 + p2(1-p2)/n2 )
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Proportion 1
Probability Succuses( X1 )=48
No.Of Observed (n1)=60
P1= X1/n1=0.8
Proportion 2
Probability Succuses(X2)=52
No.Of Observaed (n2)=132
P2= X2/n2=0.394
Margin of Error = Za/2 * Sqrt (0.004)
=0.11

PART C.
Confidence Interval for Diffrence of Proportion
CI = (p1 - p2) ± Z a/2 Sqrt(p1(1-p1)/n1 + p2(1-p2)/n2 )
Proportion 1
No. of chances( X1 )=48
No.Of Observed (n1)=60
P1= X1/n1=0.8
Proportion 2
No. of chances(X2)=52
No.Of Observed (n2)=132
P2= X2/n2=0.394
C.I = (0.8-0.394) ±Z a/2 * Sqrt( (0.8*0.2/60) + (0.394*0.606/132) )
=(0.8-0.394) ± 1.64* Sqrt(0.004)
=0.406-0.11,0.406+0.11
=[0.296,0.516]

PART D.
Given that,
sample one, x1 =48, n1 =60, p1= x1/n1=0.8
sample two, x2 =52, n2 =132, p2= x2/n2=0.394
null, Ho: p1 = p2
alternate, H1: p1 != p2
level of significance, = 0.1
from standard normal table, two tailed z /2 =
since our test is two-tailed
reject Ho, if zo < -1.645 OR if zo > 1.645
we use test statistic (z) = (p1-p2)/(p^q^(1/n1+1/n2))
zo =(0.8-0.394)/sqrt((0.521*0.479(1/60+1/132))
zo =5.22
| zo | =5.22
critical value
the value of |z | at los 0.1% is 1.645
we got |zo| =5.22 & | z | =1.645
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 5.2205 ) = 0
hence value of p0.1 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 != p2
test statistic: 5.22
critical value: -1.645 , 1.645
decision: reject Ho
p-value: 0

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