Question

Ionizing radiation is being given increasing attention as a method for preserving horticultural products. The article “The Influence of Gamma-Irradiation on the Storage Life of Red Variety Garlic” (J. Food Process. Preserv., 1983: 179-183) reports that 153 of 180 irradiated garlic bulbs were marketable (no external sprouting, rotting, or softening) 240 days after treatment, whereas only 119 of 180 untreated bulbs were marketable after this length of time. We want to test whether ionizing radiation is beneficial as far as marketability is concerned.

a) Please state the null and the alternative hypotheses H0 and Ha with appropriate symbol(s). Please define the symbol(s) properly.

b) Please report test statistic. Show your calculation.

c) Please find P-value and make a conclusion in the context of the problem (not simply whether H0 is rejected) using a 1% significance level. Please show the R code you used to find the P-value.

d) Please check all the necessary condition(s) required to perform the test above.

e) Calculate a 99% confidence interval for the change in marketable proportions after ionizing radiation. Show your calculation.

f) Calculate a 99% confidence interval for the population proportion of garlic bulbs that were marketable 240 days after the ionizing radiation treatment. Show your calculation.

Answer 1

Given that,
sample one, x1 =153, n1 =180, p1= x1/n1=0.85
sample two, x2 =119, n2 =180, p2= x2/n2=0.661
null, Ho: p1 = p2
alternate, H1: p1 > p2
level of significance, α = 0.01
from standard normal table,right tailed z α/2 =2.326
since our test is right-tailed
reject Ho, if zo > 2.326
we use test statistic (z) = (p1-p2)/√(p^q^(1/n1+1/n2))
zo =(0.85-0.661)/sqrt((0.756*0.244(1/180+1/180))
zo =4.17
| zo | =4.17
critical value
the value of |z α| at los 0.01% is 2.326
we got |zo| =4.17 & | z α | =2.326
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value: right tail - Ha : ( p > 4.1697 ) = 0.00002
hence value of p0.01 > 0.00002,here we reject Ho
ANSWERS
---------------
a.
null, Ho: p1 = p2
alternate, H1: p1 > p2
b.
test statistic: 4.17
critical value: 2.326
decision: reject Ho
c.
p-value: 0.00002
d.
we have enough evidence to support the claim that whether ionizing radiation is beneficial as far as marketability is concerned.
e.
TRADITIONAL METHOD
given that,
possible chances (x)=153
sample size(n)=180
success rate ( p )= x/n = 0.85
I.
sample proportion = 0.85
standard error = Sqrt ( (0.85*0.15) /180) )
= 0.027
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.01
from standard normal table, two tailed z α/2 =2.576
margin of error = 2.576 * 0.027
= 0.069
III.
CI = [ p ± margin of error ]
confidence interval = [0.85 ± 0.069]
= [ 0.781 , 0.919]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
possible chances (x)=153
sample size(n)=180
success rate ( p )= x/n = 0.85
CI = confidence interval
confidence interval = [ 0.85 ± 2.576 * Sqrt ( (0.85*0.15) /180) ) ]
= [0.85 - 2.576 * Sqrt ( (0.85*0.15) /180) , 0.85 + 2.576 * Sqrt ( (0.85*0.15) /180) ]
= [0.781 , 0.919]
-----------------------------------------------------------------------------------------------
interpretations:
1. We are 99% sure that the interval [ 0.781 , 0.919] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 99% of these intervals will contains the true population proportion
99% confidence interval for the change in marketable proportions after ionizing radiation.
f.
TRADITIONAL METHOD
given that,
possible chances (x)=119
sample size(n)=180
success rate ( p )= x/n = 0.661
I.
sample proportion = 0.661
standard error = Sqrt ( (0.661*0.339) /180) )
= 0.035
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.01
from standard normal table, two tailed z α/2 =2.576
margin of error = 2.576 * 0.035
= 0.091
III.
CI = [ p ± margin of error ]
confidence interval = [0.661 ± 0.091]
= [ 0.57 , 0.752]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
possible chances (x)=119
sample size(n)=180
success rate ( p )= x/n = 0.661
CI = confidence interval
confidence interval = [ 0.661 ± 2.576 * Sqrt ( (0.661*0.339) /180) ) ]
= [0.661 - 2.576 * Sqrt ( (0.661*0.339) /180) , 0.661 + 2.576 * Sqrt ( (0.661*0.339) /180) ]
= [0.57 , 0.752]
-----------------------------------------------------------------------------------------------
interpretations:
1. We are 99% sure that the interval [ 0.57 , 0.752] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 99% of these intervals will contains the true population proportion
99% confidence interval for the population proportion of garlic bulbs that were marketable 240 days after the ionizing radiation treatment.