A large company wanted to know how the average salary of their employees had changed over the last year. the results of their study had a p-value of 0.12and showed that the average salary had increased by $1,500 from $20,000 to $21,500. The results were practically significant but not statistically significant.
Answer:
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Step-by-step explanation:
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Answer:
U = 5
S = 4
1.) P(X>x) = 0.5
Prob = 1-0.5 = 0.5
We have z = 0, that is the z score with the probability of 0.5
X = u + z(s)
= 5+0*4
= 5
2.) 1-0.95 = 0.05
Z score having this probability
Z = -1.64
X = 5-1.64*4
= 5-6.56
= -1.56
3.) P(z<1.0) - p(X<x) = 0.2
0.841345-0.2 = .641345
We find the z score given this probability
Z= 0.36
X = 5+0.36*4
= 5+1.44
= 6.44
4.) P(X<x)-P(Z<-.5)
0.95 = p(X<x)-0.308538
p(X<x) = 0.308538 + 0.95
= 1.258538
There is no x value here, given that the probability is more than 1.
5. 1-0.99/2 = 0.005
We get the z score value
= -2.58
U - 5 = 5-5 = 0
-x = 0-2.58(4)
X = 10.32
Answer:
1.25
Step-by-step explanation:
