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# I have 4 more pics to upload

1 answer below Â»
Answered Same Day May 18, 2020

## Solution

Pooja answered on May 20 2020
8.142
data:    After removing outlier            a)
22    22            90% CI
19    19            mean+- t(A/2,n-1)*(sd/sqrt(n))
24    24
31    31            MEAN=    25.82
29    29            sd=    7.71
29    29            n=    22
21    21            t(a/2,n-1)=    1.7207429028
15    15
27    27            lower    22.9914814933
23    23            upper    28.6485185067
37    37
31    31            c)
30    30            90% CI
26    26            mean+- t(A/2,n-1)*(sd/sqrt(n))
16    16
26    26            MEAN=    24.7619047619
12    12            sd=    6.0490062151
23    23            n=    21
48    22            t(a/2,n-1)=    1.7247182429
22    29
29    28            lower    22.485274311
28                upper    27.0385352128
9.116
Ho:Â the 2012 mean annual expenditure on apparel and services for consumer units in the North East is equal to the national mean of \$1736
H1:Â the 2012 mean annual expenditure on apparel and services for consumer units in the North East is not equal to the national mean of \$1736
Mean= 1,922.76Â
sd= 350.90
u= 1,736.00
n= 25.00
alpha= 5%
T(a/2,n-1)
t(0.05/2,25-1)
2.064
T = (mean-u)/(sd/sqrt(n))
2.6611646086
2.6612
P-value
2*(1-P(T<|t|)
2*(1-P(T    T.DIST.2T(abs(2.6612),25-1)
0.0137
With (t=2.6612, p<5%),I reject the null hypothesis at 5% level of significance. And conclude thatÂ the 2012 mean annual expenditure on apparel and services for consumer units in the North East is not equal to the national mean of \$1736
9.154
data:    R    Sorted data            ho:    median is 7.5
8.7    2nd    5.4            h1:    median is NOT 7.5
7.5    1st    5.7
5.4    1st    5.8
13.9    2nd    6.1
7.9    2nd    7.3
5.8    1st    7.5
9.3    2nd    7.9
8.9    2nd    8.3
8.3    2nd    8.7
9.3    2nd    8.9
5.7    1st    9.3
6.1    1st    9.3            n1=    6
11.7    2nd    11.7            n2=    8
7.3    1st    13.9            u=    90
sd^2=    60
R=    8
z=     -10.5861544796
p-value=    2*(1-P(Z<=|z|)
0.000000
since p-value<5%, I reject Ho and conclude that
median is NOT 7.5
10.46
Ho:Â there is no significant difference in the average sense of direction in males and females.
H1:Â  males have a better average sense of direction as compared to female
Sample 1    Samplen 2
N=    30    30
Mean=    37.6    55.8
S=    38.5    48.3
s^2/n    49.4083    77.763
T =
= (Xbar1-Xbar2)/SQRT(S1^2/N1 + S2^2/N2)
= (37.6-55.8)/SQRT(49.4083+77.763)
-1.6139
df=
(s1^2/n1 + s2^2/n2)^2/((s1^2/n1)^2/(n1-1) + (s2^2/n2)^2/(n2-1))
(49.4083+77.763)^2/(49.4083^2/(30-1) + 77.763^2/(30-1))
186
Alpha= 1%
P-value
P(T>t)
(1-P(T<-1.6139)
T.DIST.RT(-1.6139,186)
0.945877834
With (t=-1.61, p>5%),I fail to reject the null hypothesis at 5% level of significance and conclude thatÂ there is no significant difference in the average sense of direction in males and...
SOLUTION.PDF