Question

For the following problem determine...

(a)Whether it’s a left, right or two-tail test.

(b)Whether the samples are independent or dependent.

(c)Whether we’re testing the difference of proportions or means.

(d)Which distribution we will use (and which calculator command).

(e)What the requirements would be for this test.

(f)The formula for the test statistic to be used.

(g)Perform the appropriate test.

A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.16 hours, with a standard deviation of 2.27 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.05 hours, with a standard deviation of 1.81 hours. Construct and interpret a 95% confidence interval for the mean difference in leisure time between adults with no children and adults with children.

Answer 1

Given that,
mean(x)=5.16
standard deviation , s.d1=2.27
number(n1)=40
y(mean)=4.05
standard deviation, s.d2 =1.81
number(n2)=40
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.023
since our test is two-tailed
reject Ho, if to < -2.023 OR if to > 2.023
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =5.16-4.05/sqrt((5.1529/40)+(3.2761/40))
to =2.418
| to | =2.418
critical value
the value of |t α| with min (n1-1, n2-1) i.e 39 d.f is 2.023
we got |to| = 2.41805 & | t α | = 2.023
make decision
hence value of | to | > | t α| and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 2.418 ) = 0.02
hence value of p0.05 > 0.02,here we reject Ho
ANSWERS
---------------
a.
two tailed test
b.
samples are independent
c.
t test for difference ofmeans
d.
standard normal distribution
e.
null, Ho: u1 = u2
alternate, H1: u1 != u2
f.
test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
test statistic: 2.418
critical value: -2.023 , 2.023
decision: reject Ho
p-value: 0.02
we have enough evidence to support the claim that the mean difference in leisure time between adults with no children and adults with children.
g.
TRADITIONAL METHOD
given that,
mean(x)=5.16
standard deviation , s.d1=2.27
number(n1)=40
y(mean)=4.05
standard deviation, s.d2 =1.81
number(n2)=40
I.
standard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
standard error = sqrt((5.153/40)+(3.276/40))
= 0.459
II.
margin of error = t a/2 * (standard error)
where,
t a/2 = t -table value
level of significance, α = 0.05
from standard normal table, two tailed and
value of |t α| with min (n1-1, n2-1) i.e 39 d.f is 2.023
margin of error = 2.023 * 0.459
= 0.929
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (5.16-4.05) ± 0.929 ]
= [0.181 , 2.039]
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DIRECT METHOD
given that,
mean(x)=5.16
standard deviation , s.d1=2.27
sample size, n1=40
y(mean)=4.05
standard deviation, s.d2 =1.81
sample size,n2 =40
CI = x1 - x2 ± t 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)
ta/2 = t-table value
CI = confidence interval
CI = [( 5.16-4.05) ± t a/2 * sqrt((5.153/40)+(3.276/40)]
= [ (1.11) ± t a/2 * 0.459]
= [0.181 , 2.039]
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interpretations:
1. we are 95% sure that the interval [0.181 , 2.039] contains the true population proportion
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 proportion