We have a sample of 28 data points. The sample mean is 30.0 and the sample standard deviation is 2.40. The confidence level required is 98%. Then, we calculate α by:
The confidence interval for the population mean, given the sample mean μ and the sample standard deviation σ, can be calculated as:
![CI(\mu)=\lbrack x-Z_{1-\frac{\alpha}{2}}\cdot\frac{\sigma}{\sqrt[]{n}},x+Z_{1-\frac{\alpha}{2}}\cdot\frac{\sigma}{\sqrt[]{n}}\rbrack](https://tex.z-dn.net/?f=CI%28%5Cmu%29%3D%5Clbrack%20x-Z_%7B1-%5Cfrac%7B%5Calpha%7D%7B2%7D%7D%5Ccdot%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%5B%5D%7Bn%7D%7D%2Cx%2BZ_%7B1-%5Cfrac%7B%5Calpha%7D%7B2%7D%7D%5Ccdot%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%5B%5D%7Bn%7D%7D%5Crbrack)
Where n is the sample size, and Z is the z-score for 1 - α/2. Using the known values:
![CI(\mu)=\lbrack30.0-Z_{0.99}\cdot\frac{2.40}{\sqrt[]{28}},30.0+Z_{0.99}\cdot\frac{2.40}{\sqrt[]{28}}\rbrack](https://tex.z-dn.net/?f=CI%28%5Cmu%29%3D%5Clbrack30.0-Z_%7B0.99%7D%5Ccdot%5Cfrac%7B2.40%7D%7B%5Csqrt%5B%5D%7B28%7D%7D%2C30.0%2BZ_%7B0.99%7D%5Ccdot%5Cfrac%7B2.40%7D%7B%5Csqrt%5B%5D%7B28%7D%7D%5Crbrack)
Where (from tables):
Finally, the interval at 98% confidence level is:
Odd integers are 2 apart
they are
x,x+2,x+4
sum is -75
x+x+2+x+4=-75
3x+6=-75
minus 6 both sides
3x=-81
divide both sides by 3
x=-27
x+2=-25
x+4=-23
the largest is the farthest to right on a number line
largest is -23
72/75 is 72:75 written as a fraction, it can't be simplified.
Ask yourself:How many times does 5 go into 31? Then ,how many times does 5 go into 12?
It will equal 62 R2
Ok so the equation for this is rise over run or y2-y1/x2-x1 (or y2-y1 over x2-x1)
plug your coordinates in: -4-5/0-2
-4 - 5 = -9
0-2=-2
-9/-2
But they most likely want a positive so slope should be 9/2
