To test a new method of removing nickel from fertilizers against the old method,
ID: 3314460 • Letter: T
Question
To test a new method of removing nickel from fertilizers against the old method, 50 of 100 tomato plants were randomly given fertilizer from the old method. The remaining 50 plants received fertilizer from the new method. The numbers of plants with toxic levels of nickel were recorded, and the results are below.
Method / Toxic / Non-toxic
Old / 9 / 41
New / 5 / 45
a. For each of the two methods, construct a 95% confidence interval for the proportion of plants that would have a toxic nickel level.
b. Use the hypothesis testing procedure based on the normal distribution to test to see if the new method results in a statistically significantly reduced proportion of plants with toxic nickel levels. Check all assumptions, obtain the p-value, and state your conclusion.
c. Run Fisher’s exact test and compare the results to those of (b).
d. Construct a 95% confidence interval for the difference between the two proportions.
Explanation / Answer
a.
given that,
possible chances (x)=9
sample size(n)=50
success rate ( p )= x/n = 0.18
CI = confidence interval
confidence interval = [ 0.18 ± 1.96 * Sqrt ( (0.18*0.82) /50) ) ]
= [0.18 - 1.96 * Sqrt ( (0.18*0.82) /50) , 0.18 + 1.96 * Sqrt ( (0.18*0.82) /50) ]
= [0.0735 , 0.2865]
----------------------
given that,
possible chances (x)=5
sample size(n)=50
success rate ( p )= x/n = 0.1
CI = confidence interval
confidence interval = [ 0.1 ± 1.96 * Sqrt ( (0.1*0.9) /50) ) ]
= [0.1 - 1.96 * Sqrt ( (0.1*0.9) /50) , 0.1 + 1.96 * Sqrt ( (0.1*0.9) /50) ]
= [0.0168 , 0.1832]
b.
Given that,
sample one, x1 =9, n1 =50, p1= x1/n1=0.18
sample two, x2 =5, n2 =50, p2= x2/n2=0.1
null, Ho: p1 = p2
alternate, H1: p1 > p2
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) = (p1-p2)/(p^q^(1/n1+1/n2))
zo =(0.18-0.1)/sqrt((0.14*0.86(1/50+1/50))
zo =1.153
| zo | =1.153
critical value
the value of |z | at los 0.05% is 1.645
we got |zo| =1.153 & | z | =1.645
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value: right tail - Ha : ( p > 1.1528 ) = 0.1245
hence value of p0.05 < 0.1245,here we do not reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 > p2
test statistic: 1.153
critical value: 1.645
decision: do not reject Ho
p-value: 0.1245
no evidence to support significantly reduced the proportion of plants with toxic nickel levels.
PART D.
TRADITIONAL METHOD
given that,
sample one, x1 =9, n1 =50, p1= x1/n1=0.18
sample two, x2 =5, n2 =50, p2= x2/n2=0.1
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( (0.18*0.82/50) +(0.1 * 0.9/50))
=0.0689
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
margin of error = 1.96 * 0.0689
=0.1351
III.
CI = (p1-p2) ± margin of error
confidence interval = [ (0.18-0.1) ±0.1351]
= [ -0.0551 , 0.2151]
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DIRECT METHOD
given that,
sample one, x1 =9, n1 =50, p1= x1/n1=0.18
sample two, x2 =5, n2 =50, p2= x2/n2=0.1
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 = [ (0.18-0.1) ± 1.96 * 0.0689]
= [ -0.0551 , 0.2151 ]
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interpretations:
1) we are 95% sure that the interval [ -0.0551 , 0.2151] 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, 95% of these intervals will contains the difference between
true population mean P1-P2
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