Question

In: Statistics and Probability

Evaluate, if = 0.4502 ,  = 0.4185 , n 1 = 800 , n 2 = 750,  ...

Evaluate, if

= 0.4502 ,  = 0.4185 , n 1 = 800 , n 2 = 750,  

Round the result to four digits after the point.

Evaluate, if  = 15.3 ,  = 10.2 , µ1 - µ2 = 0 , s 1= 2.1 , s 2= 2.0 , n 1 = 50, n 2 = 40

Round the result to four digits after the point.

Evaluate,

if = 0.4307 , = 0.4502 ,  = 0.4185 , n 1= 900 ,  n 2= 800.

Round the result to four digits after the point.

Evaluate, if   = 0.38 ,  = 0.39 , = 0.35 , n 1= 600 , n 2= 500.

Round the result to four digits after the point.

Evaluate, if  x 1 = 426 , x 2 = 274 , n 1= 950 ,  n 2= 800

Solutions

Expert Solution

1.
Given that,
sample one, x1 =360.16, n1 =800, p1= x1/n1=0.45
sample two, x2 =313.875, n2 =750, p2= x2/n2=0.419
null, Ho: p1 = p2
alternate, H1: p1 != p2
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = (p1-p2)/√(p^q^(1/n1+1/n2))
zo =(0.45-0.419)/sqrt((0.435*0.565(1/800+1/750))
zo =1.258
| zo | =1.258
critical value
the value of |z α| at los 0.05% is 1.96
we got |zo| =1.258 & | z α | =1.96
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 1.2581 ) = 0.2084
hence value of p0.05 < 0.2084,here we do not reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 != p2
test statistic: 1.258
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.2084
we do not have enough evidence to support the claimthat difference in proportions.
2.
Given that,
mean(x)=15.3
standard deviation , s.d1=2.1
number(n1)=50
y(mean)=10.2
standard deviation, s.d2 =2
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 =15.3-10.2/sqrt((4.41/50)+(4/40))
to =11.756
| to | =11.756
critical value
the value of |t α| with min (n1-1, n2-1) i.e 39 d.f is 2.023
we got |to| = 11.75602 & | 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 != 11.756 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 11.756
critical value: -2.023 , 2.023
decision: reject Ho
p-value: 0
we have enough evidence to support the claim that difference in means.
3.
Given that,
sample one, x1 =405.18, n1 =900, p1= x1/n1=0.45
sample two, x2 =334.8, n2 =800, p2= x2/n2=0.419
null, Ho: p1 = p2
alternate, H1: p1 != p2
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = (p1-p2)/√(p^q^(1/n1+1/n2))
zo =(0.45-0.419)/sqrt((0.435*0.565(1/900+1/800))
zo =1.316
| zo | =1.316
critical value
the value of |z α| at los 0.05% is 1.96
we got |zo| =1.316 & | z α | =1.96
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 1.3158 ) = 0.1882
hence value of p0.05 < 0.1882,here we do not reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 != p2
test statistic: 1.316
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.1882
we do not have enough evidence to support the claimthat difference in proportions.
4.
Given that,
sample one, x1 =234, n1 =600, p1= x1/n1=0.39
sample two, x2 =175, n2 =500, p2= x2/n2=0.35
null, Ho: p1 = p2
alternate, H1: p1 != p2
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = (p1-p2)/√(p^q^(1/n1+1/n2))
zo =(0.39-0.35)/sqrt((0.372*0.628(1/600+1/500))
zo =1.367
| zo | =1.367
critical value
the value of |z α| at los 0.05% is 1.96
we got |zo| =1.367 & | z α | =1.96
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 1.3668 ) = 0.1717
hence value of p0.05 < 0.1717,here we do not reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 != p2
test statistic: 1.367
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.1717
we do not have enough evidence to support the claimthat difference in proportions.
5.
Given that,
sample one, x1 =426, n1 =950, p1= x1/n1=0.448
sample two, x2 =274, n2 =800, p2= x2/n2=0.343
null, Ho: p1 = p2
alternate, H1: p1 != p2
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = (p1-p2)/√(p^q^(1/n1+1/n2))
zo =(0.448-0.343)/sqrt((0.4*0.6(1/950+1/800))
zo =4.506
| zo | =4.506
critical value
the value of |z α| at los 0.05% is 1.96
we got |zo| =4.506 & | z α | =1.96
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 4.5057 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 != p2
test statistic: 4.506
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0
we have enough evidence to support the claimthat difference in proportions.

orchestra answered 2 years ago
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