Question

1. You want to know if, on average, households have more cats or dogs. You take an SRS of 8 households and find the data below. # of cats, 2, 0, 3, 2, 0, 4, 0, 2, and the number of dogs is 1, 1, 3, 4, 0, 2, 2, 1. a) determine the populations and parameters being discussed b) determine which tool will be help us find what we need (one sample z test, one sample t test, two sample t test, one sample z interval, two sample t interval) c) Check if the conditions for this tool holds d) Whether or not the conditions hold, use the tool you chose in part (b). Use C=95% for all confidence intervals and a a=5% for all significance test. *be sure that all methods end with a sentence describing the results*

Answer 1

1.

Given that,
mean(x)=1.625
standard deviation , s.d1=1.408
number(n1)=8
y(mean)=1.75
standard deviation, s.d2 =1.198
number(n2)=8
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, α = 0.05
from standard normal table,right tailed t α/2 =1.895
since our test is right-tailed
reject Ho, if to > 1.895
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =1.625-1.75/sqrt((1.98246/8)+(1.4352/8))
to =-0.1912
| to | =0.1912
critical value
the value of |t α| with min (n1-1, n2-1) i.e 7 d.f is 1.895
we got |to| = 0.19124 & | t α | = 1.895
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value:right tail - Ha : ( p > -0.1912 ) = 0.57312
hence value of p0.05 < 0.57312,here we do not reject Ho
ANSWERS
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a.
mean(x)=1.625
standard deviation , s.d1=1.408
number(n1)=8
y(mean)=1.75
standard deviation, s.d2 =1.198
number(n2)=8
b.
sample two t test because independent variables two sets
c.
null, Ho: u1 = u2
alternate, H1: u1 > u2
test statistic: -0.1912
critical value: 1.895
decision: do not reject Ho
p-value: 0.57312
we do not have enough evidence to support the claim that if, on average, households have more cats or dogs.
d.
TRADITIONAL METHOD
given that,
mean(x)=1.625
standard deviation , s.d1=1.408
number(n1)=8
y(mean)=1.75
standard deviation, s.d2 =1.198
number(n2)=8
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((1.982/8)+(1.435/8))
= 0.654
II.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, α = 0.05
from standard normal table,right tailed and
value of |t α| with min (n1-1, n2-1) i.e 7 d.f is 1.895
margin of error = 1.895 * 0.654
= 1.239
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (1.625-1.75) ± 1.239 ]
= [-1.364 , 1.114]
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DIRECT METHOD
given that,
mean(x)=1.625
standard deviation , s.d1=1.408
sample size, n1=8
y(mean)=1.75
standard deviation, s.d2 =1.198
sample size,n2 =8
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 = [( 1.625-1.75) ± t a/2 * sqrt((1.982/8)+(1.435/8)]
= [ (-0.125) ± t a/2 * 0.654]
= [-1.364 , 1.114]
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interpretations:
1. we are 95% sure that the interval [-1.364 , 1.114] 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