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

Question

Io what extent do syntax textbooks, which analyze the structure of sentences, illustrate gender bias? A study of this question sampled sentences from 10 texts. Une part of the study examined the use of the words 'girl" "boy two wordsjwerte and the last two adut. Is the proportion or remale references that are Juvenile (girl equal to the proportion ot male references that are juvenile [boy)? Here are data trom one of the texts: "man, and "worman.' We will call the first Gender n uvenlle) Female 6047 Male 133 51 a) Find the proportion or Iuvenile references tor temales and its standard emor. Do the same tor the males. (Round your answers to three decimal places.) SEM = b] Give a 90% confidence interval tor the difference. [Do not use rounded values. Round your final answers to three decimal places.] (c) Usa a tast o fignificanca to axamina whathar tha twa proportians are aqual. ( -N. Raund your valua for z to two decimal placas and raund your P-valu to faur docimal placas.) State your cornclusiun There is not sufficient evidece to conclude that the two proportions are different. O There is sufficient evidence to condlude that the two proportions are different.

Explanation / Answer

PART A.
For females, possibile chances (x)=47
sample size(n)=60
success rate ( p )= x/n = 0.7833
sample proportion = 0.7833
standard error = Sqrt ( (0.7833*0.2167) /60) )
= 0.0532
For males, possibile chances (x)=51
sample size(n)=100
success rate ( p )= x/n = 0.51
sample proportion = 0.51
standard error = Sqrt ( (0.51*0.49) /100) )
= 0.05

PART B.
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 = [ (0.7833-0.51) ± 1.64 * 0.073]
= [ 0.1536 , 0.393 ]
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interpretations:
1) we are 90% sure that the interval [ 0.1536 , 0.393] 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

PART C.
null, Ho: p1 = p2
alternate, H1: p1 != p2
level of significance, = 0.1
from standard normal table, two tailed z /2 =1.645
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 =(0.783-0.51)/sqrt((0.613*0.388(1/60+1/100))
zo =3.436
| zo | =3.436
critical value
the value of |z | at los 0.1% is 1.645
we got |zo| =3.436 & | 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 != 3.4357 ) = 0.0006
hence value of p0.1 > 0.0006,here we reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 != p2
test statistic: 3.436
critical value: -1.645 , 1.645
decision: reject Ho
p-value: 0.0006

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