Answer:
A. 384.16
B. 2,401
C. 9,604
D. No
Step-by-step explanation:
Calculation to determine how large a sample should be taken for each desired margin of error
First step is to find σ which represent Population Standard deviation
σ=($50,000-$30,000)/4
σ=$20,000/4
σ = 5,000
Now let calculate how large a sample should be taken for each desired margin of error
Using this formula
n = (Za/2*σ/E)^2
Where,
Za/2=1-0.95/2
Za/2=0.05/2
Za/2=0.025
Z-score 0.025=1.96
Za/2=1.96
σ =5,000
E represent Desired margin of error
Let plug in the formula
a. $500
n = (1.96* 5,000/$500)^2
n=(9,800/$500)^2
n=(19.6)^2
n = 384.16
b. $200
n = (1.96*5,000/200)^2
n=(9,800/$200)^2
n=(49)^2
n = 2,401
c. $100
n = (1.96*5,000/$100)^2
n=(9,800/$100)^2
n=(98)^2
n = 9,604
Therefore how large a sample should be taken for each desired margin of error will be :
A. $500= 384.16
B. $200= 2,401
C. $100= 9,604
d.NO, Based on the information calculation i would NOT recommend trying to obtain the $100 margin of error reason been that it is highly costly compare to $500 margin of error and $200 margin of error.
Answer:
Plot and connect (5,4), (3,0) and (7,0).
Step-by-step explanation:
This is in factored form and can be graphed directly from the equation. The equation shows the x-intercepts.
The x-intercepts are found by solving for x using the zero product rule.
(x-3)=0 so x = 3
(x-7) = 0 so x = 7
The intercepts are (3,0) and (7,0). Plot the points. The vertex will occur halfway between these points. 7-3 / 2 = 4/2 = 2. This means the axis of symmetry is at x = 5 and this is the x-coordinate of the vertex too.
Substitute x = 5 into the equation and solve for y.
-(5-3)(5-7) = -(2)(-2) = 4
The vertex is (5,4). Plot it and connect the points.
Answer:
Step-by-step explanation:
Given that the observed frequencies for the outcomes as follows:
To check this we can use chi square goodness of fit test.
(Two tailed test at 5% significance level)
Assuming equally likely expected observations are found out and then chi square is calculated as (0-E)^2/E
Df = 6-1 =5
Outcome Frequency Expected frequency (Obs-exp)^2/Exp
1 36 34.83333333 0.03907496
2 30 34.83333333 0.670653907
3 41 34.83333333 1.091706539
4 40 34.83333333 0.766347687
5 23 34.83333333 4.019936204
6 39 34.83333333 0.498405104
209 209 7.086124402
p value =0.214
Since p >alpha, we accept null hypothesis
It appears that the loaded die does not behave differently than a fair die at 5% level of significance
1. add 40 to the 40 and 25
2. now you have
3. then you have
4. then square root
5.then round answer to nearest tenth which is
