3 Two identical footballs one airfilled and one heliumfilled

3. Two identical footballs, one air-filled and one helium-filled, were used outdoors on a windless day at The Ohio State University\'s athletic complex. Each football was kicked 39 times and the two footballs were alternated with each kick. The experimenter recorded the distance traveled by each ball. a. Find a 90% confidence interval for mean difference in distance traveled for the footballs. b. Test the claim that there is no difference in distance traveled for a football filled with air vs. helium

Trial   Air   Helium
1   25   25
2   23   16
3   18   25
4   16   14
5   35   23
6   15   29
7   26   25
8   24   26
9   24   22
10   28   26
11   25   12
12   19   28
13   27   28
14   25   31
15   34   22
16   26   29
17   20   23
18   22   26
19   33   35
20   29   24
21   31   31
22   27   34
23   22   39
24   29   32
25   28   14
26   29   28
27   22   30
28   31   27
29   25   33
30   20   11
31   27   26
32   26   32
33   28   30
34   32   29
35   28   30
36   25   29
37   31   29
38   28   30
39   28   26

Solution

let X denotes the distance travelled by air filled ball and Y denotes the distance travelled by helium filled ball

assume X~N(a1,b)      and Y~N(a2,b) i.e, their variances are same

we have a sample of size 39 from each of the population

so n1=sample size of X=39

n2=sample size of Y=39

let,

xbar=sample mean of X

ybar=sample mean of Y

sx²=sample variance of X

sy²=sample variance of Y

s²=((n1-1)*sx²+(n2-1)*sy²)/(n1+n2-2)

a)

so a 90% confidence interval for a1-a2 is

[xbar-ybar-s*sqrt(1/n1+1/n2)*t_n1+n2-2,0.05,xbar-ybar+s*sqrt(1/n1+1/n2)*t_n1+n2-2,0.05] where t_n1+n2-2,0.05 is the upper 0.05 point of a t distribution with degrees of freedom=n1+n2-2

now xbar=25.923    ybar=26.385   n1=39   n2=39    t_n1+n2-2,0.05=1.66515

sx²=21.968    sy²=38.611

so s²=30.2895    so s=5.504

so the 90% confidence interval is

[25.923-26.385-5.504*sqrt(2/39)*1.66515,25.923-26.385+5.504*sqrt(2/39)*1.66515]

=[-2.537,1.613] [answer]

b) here H₀: a1-a2=0      H₁:a1-a2 is not equal to 0

to test this hypothesis under 10% level of significance we have alpha=0.10 and the test statistic is

T={(xbar-ybar)-(a1-a2)}/(s*sqrt(1/n1+1/n2))~t_n1+n2-2

so under H₀ T={(xbar-ybar)}/(s*sqrt(1/n1+1/n2))~t_n1+n2-2

we reject H₀ at 10% level of significance iff

|observed value of T|> t_n1+n2-2,0.05

now |observed value of T| is 0.371<1.66515= t_n1+n2-2,0.05

hence we accept H0 and conclude that the claim \"there is no difference in distance traveled for a football filled with air vs. helium\" is true based on the given data at hand. [answer]