A random sample of size ny 16 is selected from a normal population with a mean o

Question

A random sample of size ny 16 is selected from a normal population with a mean of 75 and a standard deviation of S. A second random sample of size n2 9 is taken from another normal population with mean 70 and standard deviation 12. Let X and X2 be the two sample means. Find: The probability that X1 - X2 exceeds 4 Z= TABLE . 111 Cumulative Standard Normal Distribution (Continued) 0,00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0,09 0.0 0.500000 0503989 0.07978 0.511967 0515953 0519939 0.532922 0.527903 0531881 0.535856 0.1 0.539828 0.543795 0547758 0.551717 0.555760 0.559618 0563559 0.567495 0.571424 0.575345 02 0579260 0583166080 050954 0.594835 0.598706 0602568 0.606420 0.610261 0.614092 0.3 0.617911 0.621719 U625516 0.629300 0.633072 0636831 0.640576 0.644309 0648027 0651732 04 0655422 0.659097 0.662757 0.666402 0670031 0.673645 0,677242 0.680822 0.684386 0.687933

Explanation / Answer

Given that,
mean(x)=75
standard deviation , 1 =8
number(n1)=16
y(mean)=70
standard deviation, 2 =12
number(n2)=9
(X1-X2) =4 sample means exceed 4
null, Ho: u1 = u2
alternate, H1: 1 != u2
level of significance, = 0.02
from standard normal table, two tailed z /2 =2.326
since our test is two-tailed
reject Ho, if zo < -2.326 OR if zo > 2.326
we use test statistic (z) = (X1-X2)-(u1-u2)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
zo=(4)-(75-70)/sqrt((64/16)+(144/9))
zo =-0.2236
| zo | =0.2236
critical value
the value of |z | at los 0.02% is 2.326
we got |zo | =0.2236 & | z | =2.326
make decision
hence value of | zo | < | z | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 0.2236 ) = 0.58706
hence value of p0.02 < 0.58706,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: 1 != u2
test statistic: -0.2236
critical value: -2.326 , 2.326
decision: do not reject Ho
p-value: 0.58706
we do not have enough evidence to support the claim

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