With domestic sources of building supplies running low several years ago, roughl

Question

With domestic sources of building supplies running low several years ago, roughly 60,000 homes were built with imported Chinese drywall. According to an article federal investigators identified a strong association between chemicals in the drywall and electrical problems, and there is also strong evidence of respiratory difficulties due to the emission of hydrogen sulfide gas. An extensive examination of 52 homes found that 42 had such problems. Suppose these 52 were randomly sampled from the population of all homes having Chinese drywall. (a) Does the data provide strong evidence for concluding that more than 50% of all homes with Chinese drywall have electrical/environmental problems? Carry out a test of hypotheses using = 0.01 State the appropriate null and alternative hypotheses. Ha: P

Explanation / Answer

Given that,
possibile chances (x)=42
sample size(n)=52
success rate ( p )= x/n = 0.8077
success probability,( po )=0.5
failure probability,( qo) = 0.5
null, Ho:p<0.5  
alternate, H1: p>0.5
level of significance, = 0.01
from standard normal table,right tailed z /2 =2.33
since our test is right-tailed
reject Ho, if zo > 2.33
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.80769-0.5/(sqrt(0.25)/52)
zo =4.4376
| zo | =4.4376
critical value
the value of |z | at los 0.01% is 2.33
we got |zo| =4.438 & | z | =2.33
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: right tail - Ha : ( p > 4.4376 ) = 0
hence value of p0.01 > 0,here we reject Ho
ANSWERS
---------------
null, Ho:p<0.5
alternate, H1: p>0.5
test statistic: 4.4376
critical value: 2.33
decision: reject Ho
p-value: 0
Given that,
possibile chances (x)=42
sample size(n)=52
success rate ( p )= x/n = 0.8077
success probability,( po )=0.5
failure probability,( qo) = 0.5
null, Ho:p<0.5  
alternate, H1: p>0.5
level of significance, = 0.01
from standard normal table,right tailed z /2 =2.33
since our test is right-tailed
reject Ho, if zo > 2.33
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.80769-0.5/(sqrt(0.25)/52)
zo =4.4376
| zo | =4.4376
critical value
the value of |z | at los 0.01% is 2.33
we got |zo| =4.438 & | z | =2.33
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: right tail - Ha : ( p > 4.4376 ) = 0
hence value of p0.01 > 0,here we reject Ho
ANSWERS
---------------

test statistic: 4.44
p-value: 0

b.
given that,
possibile chances (x)=42
sample size(n)=52
success rate ( p )= x/n = 0.808
CI = confidence interval
confidence interval = [ 0.808 ± 2.33 * Sqrt ( (0.808*0.192) /52) ) ]
= [0.808 - 2.33 * Sqrt ( (0.808*0.192) /52) , 0.808 + 2.33 * Sqrt ( (0.808*0.192) /52) ]
= [0.6803 , 0.935]

Lower bound = 68.0%

c.

The probability that the test would not conclude that the proportions is more than 50%.
The probability of a type Il error as mentioned 80%.
p(0.8) = po - p + Zalpha * Sqrt( p ( 1-p)/n ) / Sqrt( p' ( 1-p')/n )
= (0.5-0.8) + 2.33 Sqrt( 0.5 (1-0.5) /52) / sqrt(0.8)(1-0.8) /52
= -2.4958
= 0.0063

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.