Answer:
Step-by-step explanation:
Hello!
You need to construct a 95% CI for the population mean of the length of engineering conferences.
The variable has a normal distribution.
The information given is:
n= 84
x[bar]= 3.94
δ= 1.28
The formula for the Confidence interval is:
x[bar]±
*(δ/n)
Lower bound(Lb): 3.698
Upper bound(Ub): 4.182
Error bound: (Ub - Lb)/2 = (4.182-3.698)/2 = 0.242
I hope it helps!
Answer:
Equal df
Step-by-step explanation:
Given that a chi square test for goodness of fit is used to examine the distribution of individuals across three categories,
Hence degree of freedom = 3-1 =2
Similarly for a chi-square test for independence is used to examine the distribution of individuals in a 2×3 matrix of categories.
Here degree of freedom = (r-1)(c-1) where r = no of rows and c =no of columns
= (2-1)(3-1) = 2
Thus we find both have equal degrees of freedom.
Independent variable. It is also called abscissa.
<span>the question is 3 variable equation solution
the three equation will be
5x+4y+7z=152
3x+2y+5z=92
2x+2y+4z=74
solving by elimination we get
x=11,y=12,z=7</span>