To what extent do syntax textbooks, which analyze the structure of sentences, il

Question

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. (Round your answers to three decimal places.)

=

SEF =

pM =

SEM =

(b) Give a 90% confidence interval for the difference. (Do not use rounded values. Round your final answers to three decimal places.)

(___,___)

(c) Use a test of significance to examine whether the two proportions are equal. (Use

pF pM. Round your value for z to two decimal places and round your P-value to four decimal places.)

=

P-value =

Explanation / Answer

a.
sample one, x1 =48, n1 =58, p1= x1/n1=0.828
sample two, x2 =54, n2 =134, p2= x2/n2=0.403

b.
TRADITIONAL METHOD
I.
standard error = sqrt( p1 * (1-p1)/n1 + p2 * (1-p2)/n2 )
where
p1, p2 = proportion of both sample observation
n1, n2 = sample size
standard error = sqrt( (0.8276*0.1724/58) +(0.403 * 0.597/134))
=0.0652
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.1
from standard normal table, two tailed z /2 =1.64
margin of error = 1.64 * 0.0652
=0.107
III.
CI = (p1-p2) ± margin of error
confidence interval = [ (0.8276-0.403) ±0.107]
= [ 0.3176 , 0.5316]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample one, x1 =48, n1 =58, p1= x1/n1=0.8276
sample two, x2 =54, n2 =134, p2= x2/n2=0.403
CI = (p1-p2) ± sqrt( p1 * (1-p1)/n1 + p2 * (1-p2)/n2 )
where,
p1, p2 = proportion of both sample observation
n1,n2 = size of both group
a = 1 - (confidence Level/100)
Za/2 = Z-table value
CI = confidence interval
CI = [ (0.8276-0.403) ± 1.64 * 0.0652]
= [ 0.3176 , 0.5316 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 90% sure that the interval [ 0.3176 , 0.5316] contains the difference between
true population proportion P1-P2
2) if a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the difference between
true population mean P1-P2

C.
null, Ho: p1 = p2
alternate, H1: p1 != p2
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = (p1-p2)/(p^q^(1/n1+1/n2))
zo =(0.828-0.403)/sqrt((0.531*0.469(1/58+1/134))
zo =5.413
| zo | =5.413
critical value
the value of |z | at los 0.05% is 1.96
we got |zo| =5.413 & | z | =1.96
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 5.4135 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 != p2
test statistic: 5.413
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0

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