Answer:
X > (whatever the symbol is) 0
Step-by-step explanation:
kinda like algebra:
x + 2 > 2
X > 0
Answer:
0.5613
Step-by-step explanation:
Here mean strain of bacteria x = sum of all numbers/ total numbers = 0.5514 %
Standard deviation s = 0.0040 %
Test statistic
G = (x_{max} -x)/s = (0.5613 - 0.5514)/ 0.0040 = 2.475
95% confidence value for outlier for n = 8 and alpha = 0.05
G_{critical} = 2.1266
so here we see that G > G_critical , so we reject that the there is no outlier . the value of 0.5613 is and outlier.
Answer:
The 90% confidence interval for the mean score of all takers of this test is between 59.92 and 64.08. The lower end is 59.92, and the upper end is 64.08.
Step-by-step explanation:
We have that to find our
level, that is the subtraction of 1 by the confidence interval divided by 2. So:
![\alpha = \frac{1-0.9}{2} = 0.05](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B1-0.9%7D%7B2%7D%20%3D%200.05)
Now, we have to find z in the Ztable as such z has a pvalue of
.
So it is z with a pvalue of
, so ![z = 1.645](https://tex.z-dn.net/?f=z%20%3D%201.645)
Now, find the margin of error M as such
![M = z*\frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%2A%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
In which
is the standard deviation of the population and n is the size of the sample.
![M = 1.645*\frac{4}{\sqrt{10}} = 2.08](https://tex.z-dn.net/?f=M%20%3D%201.645%2A%5Cfrac%7B4%7D%7B%5Csqrt%7B10%7D%7D%20%3D%202.08)
The lower end of the interval is the sample mean subtracted by M. So it is 62 - 2.08 = 59.92
The upper end of the interval is the sample mean added to M. So it is 62 + 2.08 = 64.08.
The 90% confidence interval for the mean score of all takers of this test is between 59.92 and 64.08. The lower end is 59.92, and the upper end is 64.08.
Answer:
Step-by-step explanation:
hello :
-5x+10x+3=5x+6
(-5x+10x)+3=5x+6
5x+3 = 5x+6
5x+3 -5x= 5x+6-5x
3 = 6.....(false)
no solutions in R
I found two using a little C# program:
9,10,11,12,13,14,15,16
and
18,19,20,21,22
(if you're interested in the program drop me a msg)