Answer:
3
Step-by-step explanation:
Sorry about the drawing but the line starts at the origin of the graph and since it is -x which means it has a negative slope. So the line would start at the origin going down by the slope of -1. hope this makes sense
I think the orginal expression is -27z^2-4z^2
But in simplified form it's -31z^2
Answer:
Step-by-step explanation:
Given that:
The sample mean ![\overline x = 23](https://tex.z-dn.net/?f=%5Coverline%20x%20%3D%2023)
The standard deviation
= 9
Population mean = 20
Null hypothesis:
![H_o: \mu = 20](https://tex.z-dn.net/?f=H_o%3A%20%5Cmu%20%3D%2020)
Alternative hypothesis:
![H_1 : \mu> 30](https://tex.z-dn.net/?f=H_1%20%3A%20%5Cmu%3E%2030)
(a)
When Sample size = 10
![Test \ statistics=\dfrac{\overline x - \mu }{\dfrac{\sigma }{\sqrt{n}} }](https://tex.z-dn.net/?f=Test%20%5C%20%20statistics%3D%5Cdfrac%7B%5Coverline%20x%20-%20%5Cmu%20%7D%7B%5Cdfrac%7B%5Csigma%20%7D%7B%5Csqrt%7Bn%7D%7D%20%20%7D)
![=\dfrac{23 -20}{\dfrac{9}{\sqrt{10}} }](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B23%20-20%7D%7B%5Cdfrac%7B9%7D%7B%5Csqrt%7B10%7D%7D%20%20%7D)
![=\dfrac{3 \times \sqrt{10}}{9 }](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B3%20%5Ctimes%20%5Csqrt%7B10%7D%7D%7B9%20%7D)
t = 1.0541
Degree of freedom df:
df = n -1
df = 10 - 1
df = 9
P(value) for t = 1.0541 at df = 9:
P(value) = P(Z > 1.0541)
P(value) = 1 - P(< 1.0541)
P(value) = 1 - 0.8403
P(value) = 0.1597
There is no enough evidence to infer at the 5% significance since p-value is greater than the level of significance.
(b) When sample size = 30
![Test \ statistics=\dfrac{\overline x - \mu }{\dfrac{\sigma }{\sqrt{n}} }](https://tex.z-dn.net/?f=Test%20%5C%20%20statistics%3D%5Cdfrac%7B%5Coverline%20x%20-%20%5Cmu%20%7D%7B%5Cdfrac%7B%5Csigma%20%7D%7B%5Csqrt%7Bn%7D%7D%20%20%7D)
![=\dfrac{23 -20}{\dfrac{9}{\sqrt{30}} }](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B23%20-20%7D%7B%5Cdfrac%7B9%7D%7B%5Csqrt%7B30%7D%7D%20%20%7D)
![=\dfrac{3 \times \sqrt{30}}{9 }](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B3%20%5Ctimes%20%5Csqrt%7B30%7D%7D%7B9%20%7D)
t = 1.8257
Degree of freedom df:
df = n -1
df = 30 - 1
df = 29
P(value) for t = 1.8257 at df = 29:
P(value) = P(Z > 0.9609)
P(value) = 1 - P(< 0.9609)
P(value) = 1 - 0.9609
P(value) = 0.0391
There is enough evidence to infer that the mean is greater than 20 at the 5% significance level as the p-value is less than the significance level.
(c) When sample size = 50
![Test \ statistics=\dfrac{\overline x - \mu }{\dfrac{\sigma }{\sqrt{n}} }](https://tex.z-dn.net/?f=Test%20%5C%20%20statistics%3D%5Cdfrac%7B%5Coverline%20x%20-%20%5Cmu%20%7D%7B%5Cdfrac%7B%5Csigma%20%7D%7B%5Csqrt%7Bn%7D%7D%20%20%7D)
![=\dfrac{23 -20}{\dfrac{9}{\sqrt{50}} }](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B23%20-20%7D%7B%5Cdfrac%7B9%7D%7B%5Csqrt%7B50%7D%7D%20%20%7D)
![=\dfrac{3 \times \sqrt{50}}{9 }](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B3%20%5Ctimes%20%5Csqrt%7B50%7D%7D%7B9%20%7D)
t = 2.3570
Degree of freedom df:
df = n -1
df = 50 - 1
df = 49
P(value) for t = 2.3570 at df = 49:
P(value) = P(Z > 0.9888)
P(value) = 1 - P(< 0.9888)
P(value) = 1 - 0.9888
P(value) = 0.0112
There is enough evidence to infer that the mean is greater than 20 at the 5% significance level as the p-value is less than the significance level.
It moves the graph 5 steps downwards on the y axis.