On the island of Lilliput, a sample random sample of 200 people reveals that 180
ID: 3220717 • Letter: O
Question
On the island of Lilliput, a sample random sample of 200 people reveals that 180 of them prefer to open their egg on the small end. On the island of Blefuscu, a sample random sample of 240 people reveals that 204 of them prefer to open their egg on the small end. 1.)Assuming that Lilliputians and Blefuscuans are equally likely to open their egg on the small end, what is the probability of selecting samples of 200 Lilliputians and 240 Blefuscuans with sample proportions as different as ours were (two-sided p-value, to four decimal places): The ideal (daytime) noise-level for hospitals is 45 decibels with a standard deviation of 10 db. A simple random sample of 81 hospitals at a moment during the day gives a mean noise level of 47 db. Assume that the standard deviation of noise level is really 10 db. 2.)Assuming that the average noise level of hospitals is what it's supposed to be, what is the probability of a sample of 81 hospitals producing an average as high as our sample's (p-value)?
Explanation / Answer
Given that,
sample one, x1 =180, n1 =200, p1= x1/n1=0.9
sample two, x2 =204, n2 =240, p2= x2/n2=0.85
null, Ho: p1 = p2
alternate, H1: p1 != p2
level of significance, = 0.05
from standard normal table, two tailed z /2 =
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 =(0.9-0.85)/sqrt((0.873*0.127(1/200+1/240))
zo =1.567
| zo | =1.567
critical value
the value of |z | at los 0.05% is 1.96
we got |zo| =1.567 & | z | =1.96
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 1.567 ) = 0.1171
hence value of p0.05 < 0.1171,here we do not reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 != p2
test statistic: 1.567
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.1171
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