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Answered Same Day May 18, 2020

Solution

Pooja answered on May 20 2020
123 Votes
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: 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: 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= 1,922.76 
    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 that 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: there is no significant difference in the average sense of direction in males and females.
    H1:  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 that there is no significant difference in the average sense of direction in males and...
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