. Radon Gas Abatement: The installation of a radon abatement device is recommend

Question

. Radon Gas Abatement: The installation of a radon abatement device is recommended in any home where the mean radon concentration is 4.0 picocuries per liter (pCi/L) or more, because it is thought that long-term exposure to sufficiently high doses of radon can increase the risk of cancer. Seventy- five measurements are made in a home. The average concentration was 3.72 pCi/L and the sample standard deviation was 1.93 pCi/L 4. The home inspector who performed the test say that since the average measurement is less than 4.0, radon abatement is not necessary. Explain why this is wrong. 5. Because of health concerns, radon abatement is recommended whenever it is plausible that mean radon concentration may be 4.0 pCi/L or more. State the appropriate null alternative hypotheses for determining whether radon abatement is appropriate. 6. Compute the p-value. Would you recommend radon abatement? Explain 7. Compute a 95% upper confidence bound on the mean radon concentration. Is this consistent with your conclusion in part c?

Explanation / Answer

Q4.
since the average value is expecte mean of the distribution, values in the sample is diffrent to the
average, and it is not necessarily true data values are less than 4 for the currentsample

Q5.
null, Ho: >=4
alternate, H1: <4

Q6.
Given that,
population mean(u)=4
sample mean, x =3.72
standard deviation, s =1.93
number (n)=75
level of significance, = 0.05
from standard normal table,left tailed t /2 =1.666
since our test is left-tailed
reject Ho, if to < -1.666
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =3.72-4/(1.93/sqrt(75))
to =-1.2564
| to | =1.2564
critical value
the value of |t | with n-1 = 74 d.f is 1.666
we got |to| =1.2564 & | t | =1.666
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :left tail - Ha : ( p < -1.2564 ) = 0.10646
hence value of p0.05 < 0.10646,here we do not reject Ho
ANSWERS
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null, Ho: >=4
alternate, H1: <4
test statistic: -1.2564
critical value: -1.666
decision: do not reject Ho
p-value: 0.10646

we support the claim

Q7.
given that,
sample mean, x =4
standard deviation, s =3.72
sample size, n =75
level of significance, = 0.05
from standard normal table,left tailed value of |t /2| with n-1 = 74 d.f is 1.666
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 4 ± t a/2 ( 3.72/ Sqrt ( 75) ]
= [ 4-(1.666 * 0.43) , 4+(1.666 * 0.43) ]
= [ 3.284 , 4.716 ]
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interpretations:
1) we are 95% sure that the interval [ 3.284 , 4.716 ] contains the true population mean
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 true population mean

we have the similar decision with confidence, as zero is not lies in the interval

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