A1)
a) The 98% Ci for the difference between population means is :
( x1 − x 2 ) ± Z 0.01 × S ×
!
1 1
+
n1 n2
Where,
x1 = 0.52, n1 = 10, s1 = 0.02
x 2 = 0.5, n2 = 15, s 2 = 0.01
Z 0.01 = 2.3263
S=
!
2
(n1 − 1) s12 + (n2 − 1) s 2
n1 + n2 − 2
Using all the values , the confidence interval is given by:
(0.006, 0.034)
b)
Let ! µ1 denote the mean reaction time of non players of video games in the population and
! µ 2 denote the mean reaction time of players of video games in the population.
Null hypothesis, Ho: ! µ1 =! µ 2
Alternative hypothesis, H1: ! µ1 >! µ 2
( x1 − x 2 ) /{S ×
The test statistic :Z= !
1
1
+ }
n1 n2
Where,
x1 = 0.52, n1 = 10, s1 = 0.02
x 2 = 0.5, n2 = 15, s 2 = 0.01
S=
!
2
(n1 − 1) s12 + (n2 − 1) s 2
n1 + n2 − 2
Thus Z=3.3226
The tabulated value at 2.5 % LOS is 1.9596
Thus HO is rejected as the calculated value is greater than tabulated value. Thus the
researcher can conclude his hypothesis.
A2)
a)The 97.5% CI for the difference in the population proportion is given by:
ˆ ˆ
( p1 − p 2 ) ± Z 0.0125 ×
!
ˆ
ˆ
ˆ
ˆ
p1 × (1 − p1 ) p 2 × (1 − p 2 )
+
n1
n2
ˆ
ˆ
Where, ! p1 is the sample proportion of Americans and ! p 2 is the sample proportion of
Canadians.
550
= 0.55
1000
225
ˆ
p2 =
= 0.45
500
n1 = 1000
ˆ
p1 =
n2 = 500
!
Z 0.0125 = 2.2414
Thus the CI is (.0389,0.1610)
b)
Let p1 and p2 be the population proportion for Americans and Canadians
respectively/
H0:p1=p2
HI:p1! ≠ p2
ˆ ˆ
( p1 − p 2 ) /{
Test stat: Z=!
ˆ
ˆ
ˆ
ˆ
p1 × (1 − p1 ) p 2 × (1 − p 2 )
+
}
n1
n2
Using all the values from part a), Z =3.6698
The calculated value is greater than tabulated value so we reject the null hypothesis.
A3)
Here we use the paired t test .
Let xi’s denote the number of hours before program and yi’s denote the number of
hours after program. Then di’s denote the differences between xi’s and yi’s.
d
The test statistic is t=! S d / n
~ t n −1
Where ,
d =
∑d
!
Sd =
i
i
n
∑ (d
i
− d )2
i
n −1
Thus using the given values , t=-1.04
The tabulated value is 1.89
Since abs t is less than tab t , we do no have sufficient evidence to reject the
null hypothesis. So we say there is no significant difference in the before and after
program mean hours.
A4)
Here we would set up a chi square test
SEASON
OBS FREQ
EXPECTED FREQ
!
Fall
80
70
1.4285
Winter
40
50
2
Spring
70
60
1.667
Summer
10
20
( o i − ei ) 2 / ei
5
The expected frequencies are calculated using the given percentage for each season and
expressing for 200 Like for Fall it is 35% and 35% of 200 is 70 and likewise for others. Now the
test statistic is :
( o i − ei ) 2
2
χ =∑
~ χ n −1
ei
i
!
2
So from the above table , the calculated value is 10.0955 (1.4285+2+1.667+5)
Tabulated value is 7.814.Since calc value is greater than the tabulated one , we conclude that
observed data contradicted the hypothesis.
A5)
We set up the contingency tables as below:
OBSERVED VALUES
OPINION
NEUTRA
OPPOSED L
UNDER 25
IN
FAVOUR
TOTAL
OVER 55
5
10
20
10
30
10
50
15
10
5
30
30
AGE 35-55
5
45
25
100
!
!
EXPECTED VALUES
OPINION
OPPOSED
UNDER 25
AGE
NEUTRAL
IN FAVOUR
TOTAL
6
22,5
12,5
50
13,5
7,5
30
30
!
!
!
20
9
OVER 55
5
15
35-55
9
45
25
100
Expected values are calculated using{( row total *column total)/grand total}The test
χ2 = ∑
( o i − ei ) 2
ei
with (r-1)*(c-1) df where is the # of rows and c is the #
i
statistic is given as: !
of columns. Using the values from above and the formula, we get calculated value as
17.352.The tabulated value at 1% LOS is: 13.2767.Since calculated value is greater than the
tabulated value, we reject the hypothesis that there is no relation between age and opinion
A6)
a) The 95% CI for the population variance is given by:
!
(n − 1) S 2
(n − 1) S 2
>σ2 >
χ (2 −α / 2 )
χ (2α / 2 )
1
Here ,
α = 0.05
S 2 = 25
n = 16
2
χ 0.975,15 = 6.262
2
χ 0.025,15 = 27.4884
!
Using all the values, the CI is (13.64211,59.885)
The CI for standard dev is(3.6935,7.7385)
b)
!
HO : σ 2 = 100
2
! H 1 : σ < 100
(n − 1) S 2
~ χ (2n −1)
σ2
Test statistic:!
Thus calculated value is: 3.75
Tabulated value is 25.Since Calculated value is les than tabulated value we don’t
have enough evidence to reject HO
A7) Let !
µ1 , µ 2 , µ 3 denote mean times to relieve pain for Brand A,B,C respectively
H0: !
µ1 = µ 2 = µ 3
H1:Atleast 2 are different
We run one way ANOVA
Anova: Single Factor
SUMMARY
Groups
Count
Sum
Average
Variance
BRAND A
4
51
12.75 4.916667
BRAND B
4
81
20.25
18.25
BRAND C
4
76
19
34
ANOVA
Source of
Variation
SS
Between Groups
129.1667
Within Groups
Total
171.5
300.6667
df
MS
F
P-value
F crit
2 64.58333 3.389213 0.079947 5.714705
9 19.05556
11
!
From the table since the p value is greater than 0.025, we do not have sufficient evidence to
reject the null hypothesis, thus the mean time is not statistically significant for 3 brands
!
