Question

1. The time to solve a puzzle is being measured for two groups, elderly (over 70) and kids (under 10). For the 120 adults tested it took an average time of 24 minutes with a standard deviation of 4 minutes to solve the puzzle. For the sample of 88 kids it took an average time of 22 minutes with a standard deviation of 5 minutes to solve the puzzle.

a) At the .05 level is there a difference in the average time between kids and adults to solve the puzzle? (p = .002)

b) Based on a) how do explain the 2 minute difference in average of the two groups?

(11) 2. A car has a gas mileage listing of 28 miles per gallon. You take the car out on 56 trips and find the average gas mileage to be 25 with a standard deviation of 6 mpg.

a) At the .01 level is the car getting gas mileage less than expected? (p = 0)

b) True or false: natural variability is the reason for the change in gas mileage

Answer 1

1.

a.
Given that,
mean(x)=24
standard deviation , s.d1=4
number(n1)=120
y(mean)=22
standard deviation, s.d2 =5
number(n2)=88
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =1.988
since our test is two-tailed
reject Ho, if to < -1.988 OR if to > 1.988
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =24-22/sqrt((16/120)+(25/88))
to =3.0956
| to | =3.0956
critical value
the value of |t α| with min (n1-1, n2-1) i.e 87 d.f is 1.988
we got |to| = 3.09557 & | t α | = 1.988
make decision
hence value of | to | > | t α| and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 3.0956 ) = 0.003
hence value of p0.05 > 0.003,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 3.0956
critical value: -1.988 , 1.988
decision: reject Ho
p-value: 0.003
we have enough evidence to support the claim that difference in the average time between kids and adults to solve the puzzle
b.
yes,
the 2 minute difference in average of the two groups,
2.
a.
Given that,
population mean(u)=28
sample mean, x =25
standard deviation, s =6
number (n)=56
null, Ho: μ=28
alternate, H1: μ<28
level of significance, α = 0.01
from standard normal table,left tailed t α/2 =2.396
since our test is left-tailed
reject Ho, if to < -2.396
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =25-28/(6/sqrt(56))
to =-3.7417
| to | =3.7417
critical value
the value of |t α| with n-1 = 55 d.f is 2.396
we got |to| =3.7417 & | t α | =2.396
make decision
hence value of | to | > | t α| and here we reject Ho
p-value :left tail - Ha : ( p < -3.7417 ) = 0.00022
hence value of p0.01 > 0.00022,here we reject Ho
ANSWERS
---------------
null, Ho: μ=28
alternate, H1: μ<28
test statistic: -3.7417
critical value: -2.396
decision: reject Ho
p-value: 0.00022
we have enough evidence to support the claim that the car getting gas mileage less than expected.
b.
true, because it reject the null hypothesis
natural variability is the reason for the change in gas mileage.