A18 in appends tion of coffee of the 41 males. Of the sample of the 41 males, th
ID: 3324700 • Letter: A
Question
A18 in appends tion of coffee of the 41 males. Of the sample of the 41 males, the sample mean is 25.7 gallons and the sample standard deviation s 3.8 gallons. Assume the data is normally distributed and the sample is randomly seled. Assuming -os, is there enough evidence to support a claim at least 28 gallons? (12 points) 7.32. at the mean annount or consumptona 3. Assuming = 0.01, is there enough evidence to support a claim that the standard deviation of the #5 number ofservingsis3 gallons? (12points)Explanation / Answer
2.
Given that,
population mean(u)=28
sample mean, x =25.7
standard deviation, s =3.8
number (n)=41
null, Ho: =28
alternate, H1: !=28
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.021
since our test is two-tailed
reject Ho, if to < -2.021 OR if to > 2.021
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =25.7-28/(3.8/sqrt(41))
to =-3.876
| to | =3.876
critical value
the value of |t | with n-1 = 40 d.f is 2.021
we got |to| =3.876 & | t | =2.021
make decision
hence value of | to | > | t | and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -3.8756 ) = 0.0004
hence value of p0.05 > 0.0004,here we reject Ho
ANSWERS
---------------
null, Ho: =28
alternate, H1: !=28
test statistic: -3.876
critical value: -2.021 , 2.021
decision: reject Ho
p-value: 0.0004
we have enough evidence to support the claim that mean amount of consumption is atleast 28 gallons
3.
Given that,
population standard deviation ()=3
sample standard deviation (s) =3.8
sample size (n) = 41
we calculate,
population variance (^2) =9
sample variance (s^2)=14.44
null, Ho: =3
alternate, H1 : !=3
level of significance, = 0.01
from standard normal table, two tailed ^2 /2 =63.691
since our test is two-tailed
reject Ho, if ^2 o < - OR if ^2 o > 63.691
we use test statistic chisquare ^2 =(n-1)*s^2/o^2
^2 cal=(41 - 1 ) * 14.44 / 9 = 40*14.44/9 = 64.178
| ^2 cal | =64.178
critical value
the value of |^2 | at los 0.01 with d.f (n-1)=40 is 63.691
we got | ^2| =64.178 & | ^2 | =63.691
make decision
hence value of | ^2 cal | > | ^2 | and here we reject Ho
^2 p_value =0.009
ANSWERS
---------------
null, Ho: =3
alternate, H1 : !=3
test statistic: 64.178
critical value: -63.691 , 63.691
p-value:0.009
decision: reject Ho
we have enough evidence to support that standard deviation of the number of servings is 3 gallons
4.
Given that,
population mean(u)=20.2
standard deviation, =4.8
sample mean, x =21.6
number (n)=92
null, Ho: =20.2
alternate, H1: >20.2
level of significance, = 0.04
from standard normal table,right tailed z /2 =1.751
since our test is right-tailed
reject Ho, if zo > 1.751
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 21.6-20.2/(4.8/sqrt(92)
zo = 2.798
| zo | = 2.798
critical value
the value of |z | at los 4% is 1.751
we got |zo| =2.798 & | z | = 1.751
make decision
hence value of | zo | > | z | and here we reject Ho
p-value : right tail - ha : ( p > 2.798 ) = 0.003
hence value of p0.04 > 0.003, here we reject Ho
ANSWERS
---------------
null, Ho: =20.2
alternate, H1: >20.2
test statistic: 2.798
critical value: 1.751
decision: reject Ho
p-value: 0.003
we have enough evidence to support the claim that mean amount of consumption is more than 20.2 gallons
5.
Given that,
value of r =0.623
number (n)=92
null, Ho: =0
alternate, H1: !=0
level of significance, = 0.01
from standard normal table, two tailed t /2 =2.632
since our test is two-tailed
reject Ho, if to < -2.632 OR if to > 2.632
we use test statistic (t) = r / sqrt(1-r^2/(n-2))
to=0.623/(sqrt( ( 1-0.623^2 )/(92-2) )
to =7.556
|to | =7.556
critical value
the value of |t | at los 0.01% is 2.632
we got |to| =7.556 & | t | =2.632
make decision
hence value of | to | > | t | and here we reject Ho
ANSWERS
---------------
null, Ho: =0
alternate, H1: !=0
test statistic: 7.556
critical value: -2.632 , 2.632
decision: reject Ho