Normal Curve Areas 0120 055m .069 .0675 tN910 .1026 217 -1255 406 14A3 1915 1950

Question

Normal Curve Areas 0120 055m .069 .0675 tN910 .1026 217 -1255 406 14A3 1915 1950 808 1985 2019 41 1517 .2190 2517 2580 2910 .3212 2794 .2823 ,2852 ,3051 3133 323 3508 ,3315 1.0 3554 3749 3888 4082 A251 1.2 ,4015 .4147 ,4131 A177 1.4 A515 A599 A573 1.9 .4713 .4719 .4726 .4732 A738 4744 4750 A756 4761 ,4767 2.0 .4772 .4778 .4783 .4788 4793 .4798 -4803 .4808 4812 4817 .4834 Asi A922 4925 A945 4878 2.2 ,4913 ,4916 ,4918 .4931 2.5 4971 4972 4978 4984 4965 A966 .4967 .4976 A982 4983 4907 .4981 4989 4989 3.1 49903 4990% .49910 .49913 .49916 49918 .49921 A9924 A9926 .48829 N 49931 .49934 49936 49938 .49940 .49942 49944 .49946 .49948 .49950 49962 49964 .49965 8.4 49966 A9968 .49969 49970 49971 49972 49973 A9974 49975 49976 49977 49978 49978 49979 49980 .49981 .49981 .49982 .49983 49983 49984 .49985 .49985 .49986 ,49986 .49987 .49987 .49988 .4998s .49g89 49989 49990 ,49990 49990 49991 .49991 A992 A992 A9992 .4992 .49994 .49994 .49994 A9994 .49095 49993 4993 .49993 g .49995 .49995 .49996 .49996 .49996 49996 49996 49996 49997 e Abridged from Tablelof A Hald Statistical Tables and Formulas (New York: Waey), 1952. Reproduced by permission of A. Hald.

Explanation / Answer

Q1.
a.
Confidence Interval For Proportion
CI = p ± Z a/2 Sqrt(p*(1-p)/n)))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
No. of success(x)=62
Sample Size(n)=93
Sample proportion = x/n =0.667
Confidence Interval = [ 0.667 ±Z a/2 ( Sqrt ( 0.667*0.333) /93)]
= [ 0.667 - 2.576* Sqrt(0.002) , 0.667 + 2.58* Sqrt(0.002) ]
= [ 0.541,0.793]
b.
Given that,
possibile chances (x)=62
sample size(n)=93
success rate ( p )= x/n = 0.67
success probability,( po )=0.6
failure probability,( qo) = 0.4
null, Ho:p<0.6
alternate, H1: p>0.6
level of significance, = 0.1
from standard normal table,right tailed z /2 =1.28
since our test is right-tailed
reject Ho, if zo > 1.28
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.66667-0.6/(sqrt(0.24)/93)
zo =1.31
| zo | =1.31
critical value
the value of |z | at los 0.1% is 1.28
we got |zo| =1.312 & | z | =1.28
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: right tail - Ha : ( p > 1.31233 ) = 0.0947
hence value of p0.1 > 0.0947,here we reject Ho
ANSWERS
---------------
null, Ho:p=0.6
alternate, H1: p>0.6
test statistic: 1.31
critical value: 1.28
decision: reject Ho
p-value: 0.0947
we have evidence that strikes is greater than 0.60
c.
Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)
Z a/2 at 0.05 is = 1.96
Sample Proportion = 0.667
ME = 0.01
n = ( 1.96 / 0.01 )^2 * 0.667*0.333
= 8532.616 ~ 8533

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