Question

16. A sociologist surveys 50 randomly selected citizens in each of two countries to compare the mean number of hours of volunteer work done by adults in each. Among the 50 inhabitants of Lilliput, the mean hours of volunteer work per year was 52, with standard deviation 11.8. Among the 50 inhabitants of Blefuscu, the mean number of hours of volunteer work per year was 37, with standard deviation 7.2. Construct the 99% confidence interval for the difference in mean number of hours volunteered by all residents of Lilliput and the mean number of hours volunteered by all residents of Blefuscu. Test, at the 1% level of significance, the claim that the mean number of hours volunteered by all residents of Lilliput is more than ten hours greater than the mean number of hours volunteered by all residents of Blefuscu. Compute the observed significance of the test in part (b).

Answer 1

Q1. confidence interval;
given that,
mean(x)=52
standard deviation , σ1 =11.8
number(n1)=50
y(mean)=37
standard deviation, σ2 =7.2
number(n2)=50
level of significance, α = 0.01
from standard normal table, two tailed z α/2 =2.576
CI = x1 - x2 ± Z a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
Za/2 = Z-table value
CI = confidence interval
CI = [ ( 52-37) ±Z a/2 * Sqrt( 139.24/50+51.84/50)]
= [ (15) ± Z a/2 * Sqrt( 3.8216) ]
= [ (15) ± 2.576 * Sqrt( 3.8216) ]
= [9.9642 , 20.0358]
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interpretations:
1. we are 99% sure that the interval [9.9642 , 20.0358] contains the difference between
true population mean U1 - U2
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 difference between
true population mean U1 - U2
3. Since this Cl does contain a zero we can conclude at 0.01 true mean
difference is zero
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Q2.
Given that,
mean(x)=52
standard deviation , σ1 =11.8
number(n1)=50
y(mean)=37
standard deviation, σ2 =7.2
number(n2)=50
null, Ho: u1 - u2 = 10
alternate, H1: μ1 - u2 > 10
level of significance, alpha = 0.01
from standard normal table,right tailed z alpha/2 =2.326
since our test is right-tailed
reject Ho, if zo > 2.326
we use test statistic (z) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
zo=((52-37)-10)/sqrt((139.24/50)+(51.84/50))
zo =2.557
| zo | =2.557
critical value
the value of |z alpha| at los 0.01% is 2.326
we got |zo | =2.5573 & | z alpha | =2.326
make decision
hence value of | zo | > | z alpha| and here we reject Ho
p-value : right tail - Ha :( p > 2.557) = 0.005279          
hence value of p0.01 > 0,here we reject Ho
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null, Ho: u1 - u2 = 10
alternate, H1: μ1 - u2 > 10
test statistic: 2.557
critical value: 2.326
decision: reject Ho
p-value: 0.005279
evidence that mean number of hours volunteered by all residents of lilliput is more than
ten hours greater than the mean number of hours volunteered by all residents of blefuscu