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 1 year ago

Next > < Previous

Related Solutions

Evaluate if the followings are Cauchy sequences or not. (a) an= (-1)n (b) an= (-1)n/n (c)...

Evaluate if the followings are Cauchy sequences or not. (a) an= (-1)n (b) an= (-1)n/n (c) an = n/(n+1) (d) an = (cos n)/n

A recent study asked 800 women and 750 men if they expected to be promoted at...

A recent study asked 800 women and 750 men if they expected to be promoted at their companies in the next 12 months. 160 women and 190 men said yes. At the .05 level, is there a significant difference between women and men in their expectations of receiving a promotion? a. What is the pooled proportion? b. what is the test statistic and p-value? c. is gender difference significant? d. Calculate the 95% confidence interval for the difference

Show that (a)Sn=<(1 2),(1 3),……(1 n)>. (b)Sn=<(1 2),(2 3),……(n-1 n)> (c)Sn=<(1 2),(1 2 …… n-1 n)>

On August 1, 2019, Dillard sold 800 bundles super shaver complete. Each bundle sold for $750...

On August 1, 2019, Dillard sold 800 bundles super shaver complete. Each bundle sold for $750 a bundle consisting of 1) super shaver with a retail cost of $100 and a 24 month subscription to super shaver supplies. The supplies if sold separately would sell for $30 per month. Dillard recorded sales revenue of $600,000 in 2019 related to these bundles. All costs related to the super shaver and the shipments of supplies were expensed properly. Please prepare all appropriate...

Which codes add 1 to integer n? 1) n=n+1; 2) n++; 3)++n; 4) n+=1; 5) n=--n+2

R= Ro(1/2)n n= number of half lifetimes= t/t1/2 (a)n= 2/2= 1 R= Ro(1/2)n R=3000(1/2)1 R= 1500...

R= Ro(1/2)n n= number of half lifetimes= t/t1/2 (a)n= 2/2= 1 R= Ro(1/2)n R=3000(1/2)1 R= 1500 counts/sec (b)n= 6/2= 3 R= Ro(1/2)n R=3000(1/2)3 R= 375 counts/sec (c) n= 10/2= 5 R= Ro(1/2)n R=3000(1/2)5 R= 93.75 counts/sec (d) n= 20/2= 10 R= Ro(1/2)n R=3000(1/2)10 R= 2.93 counts/sec What is the mean life of this nucleus? f. Suppose that the Geiger counter detects 10% of all the radioactive decays. What is the total number of radioactive nuclei at time t = 0?...

def annoying_factorial(n): if n == 0 or n == 1: return 1 if n == 2:...

def annoying_factorial(n): if n == 0 or n == 1: return 1 if n == 2: return 2 if n == 3: return 6 if n == 4: return 4 * annoying_factorial(3) if n == 5: return 5 * annoying_factorial(4) if n == 6: return 6 * annoying_factorial(5) else: return n * annoying_factorial(n-1) def annoying_fibonacci(n): if n==0: return 0 if n==1: return 1 if n==2: return 1 if n==3: return 2 if n==4: return annoying_fibonacci(4-1)+annoying_fibonacci(4-2) if n==5: return annoying_fibonacci(5-1)+annoying_fibonacci(5-2) if...

Prove via induction the following properties of Pascal’s Triangle: •P(n,2)=(n(n-1))/2 • P(n+m+1,n) = P(n+m,n)+P(n+m−1,n−1)+P(n+m−2,n−2)+···+P(m,0) for all...

Prove via induction the following properties of Pascal’s Triangle: •P(n,2)=(n(n-1))/2 • P(n+m+1,n) = P(n+m,n)+P(n+m−1,n−1)+P(n+m−2,n−2)+···+P(m,0) for all m ≥ 0

Define the following function f(n) = 5(2^n)-(2^(n-1)), n ≥ 1. Write a recursive definition for the...

Define the following function f(n) = 5(2^n)-(2^(n-1)), n ≥ 1. Write a recursive definition for the function f(n)? Consider the following recurrence: an= 2an-1 + 3 (where a1 = 1). Compute the values of an for n ≤ 5. Find a solution for the recurrence definition and validate its correctness. Consider the following recurrence: an=2an-1 +an-1an-2 (where a1 = 1). Compute the values of an for n ≤ 5.

Evaluate the following integral using the Trapezoid Rule with n = 4 where a = 2....

Evaluate the following integral using the Trapezoid Rule with n = 4 where a = 2. 2 ∫ rootsquare (x^3 + a) dx 0 Round your answer to 1 decimal place.

Subjects