a) The New York School Board records the graduation rates for two teachers. In M
ID: 3330276 • Letter: A
Question
a) The New York School Board records the graduation rates for two teachers. In Ms. Fortune’s class, 37 students graduated out of 55. In Miss Information’s class, 78 students graduated out of 100. The School board fires Miss Information, and not Ms. Fortune. Explain how a lurking or confounding variable might explain this choice. Make up some data to illustrate your answer.
b) Farmer Fischer is interested in growing a field of beautiful sunflowers. He decides to fertilize one field, but not the other. In the first field (the fertilized one), he finds that the flowers have a mean height of 4.2 feet, with a standard deviation of 1.2 feet. In the second field, he finds that the flowers have a mean height of 3.8 feet with a standard deviation of 2.3 feet. Make a convincing argument that the fertilizer did or did not help flower growth.
c) Wheeler Dancer offers you the opportunity to play the following game: He will roll one die. If it is a 6, you get $1. If not, he rolls again. If it is a 6, you get $2. If not, he rolls again, and if that is a 6, you get $3, etc. The game continues until a 6 is rolled. Unfortunately, the game costs $5 to play. Do you accept Mr. McDealer’s offer? (Justify your answer)
Explanation / Answer
a.
Given that,
sample one, x1 =37, n1 =55, p1= x1/n1=0.673
sample two, x2 =78, n2 =100, p2= x2/n2=0.78
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 =(0.673-0.78)/sqrt((0.742*0.258(1/55+1/100))
zo =-1.46
| zo | =1.46
critical value
the value of |z | at los 0.05% is 1.96
we got |zo| =1.46 & | 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 != -1.4604 ) = 0.1442
hence value of p0.05 < 0.1442,here we do not reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 != p2
test statistic: -1.46
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.1442
we do not have enough evidence to support the claim
b.
Given that,
mean(x)=4.2
standard deviation , 1 =1.2
number(n1)=1
y(mean)=3.8
standard deviation, 2 =2.3
number(n2)=1
null, Ho: u1 = u2
alternate, H1: 1 > u2
level of significance, = 0.05
from standard normal table,right tailed z /2 =1.645
since our test is right-tailed
reject Ho, if zo > 1.645
we use test statistic (z) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
zo=4.2-3.8/sqrt((1.44/1)+(5.29/1))
zo =0.15
| zo | =0.15
critical value
the value of |z | at los 0.05% is 1.645
we got |zo | =0.154 & | z | =1.645
make decision
hence value of | zo | < | z | and here we do not reject Ho
p-value: right tail -Ha : ( p > 0.15 ) = 0.43873
hence value of p0.05 < 0.43873,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: 1 > u2
test statistic: 0.15
critical value: 1.645
decision: do not reject Ho
p-value: 0.43873
we do not have enough evidence to support the claim
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