Question

Question 1. The numbers 5, 8, 10, 7, 10, and 14 gather from a simple random sample of a population. Determine/Calculate the point estimate of the population mean (or worded another way, what is the sample mean, x-bar,?). Determine/Calculate the point estimate of the population standard deviation (or worded another way, what is the sample standard deviation, s,?).

Question 2. A population has a mean of 200 dollars earned a day and a standard deviation of 50 dollars a day. Suppose a simple random sample was taken of 100 people (note: n > 30 so we can assume standard distribution) and we used x-bar (sample mean) to estimate mu (population mean). What is the probability that the sample mean will be within +/- 5 of the population mean? What is the probability that the sample mean will be within +/- 10 of the population mean?

Question 3. A population proportion was determined to be 0.40. A simple random sample was taken of 200 people will be taken and the sample proportion, p-bar, will be used to estimate the population proportion (which we already know is 0.40). What is the probability that the sample proportion (p-bar) will be within +/- 0.03 of the population proportion? What is the probability that the sample proportion (p-bar) will be within +/- 0.05 of the population proportion?

Answer 1

1)

X	(X - X̄)²
5	16.00
8	1.00
10	1.00
7	4.00
10	1.00
14	25.00

	X	(X - X̄)²
total sum	54	48.00
n	6	6

mean =    ΣX/n =    54.000   /   6   =   9.0000
                            
sample std dev =   √ [ Σ(X - X̄)²/(n-1)] =   √   (48/5)   =       3.0984

..............

2)

µ =    200                                  
σ =    50                                  
n=   100     

for +-5
we need to calculate probability for ,                                      
195   ≤ X ≤    205                              
X1 =    195   ,    X2 =   205                      
                                      
Z1 =   (X1 - µ )/(σ/√n) = (   195   -   200   ) / (   50   / √   100   ) =   -1.00
Z2 =   (X2 - µ )/(σ/√n) = (   205   -   200   ) / (   50   / √   100   ) =   1.00
                                      
P (   195   < X <    205   ) =    P (    -1.0   < Z <    1.0   )   
                                      
= P ( Z <    1.00   ) - P ( Z <   -1.00   ) =    0.84134   -    0.15866   =    0.6827  
excel formula for probability from z score is =NORMSDIST(Z)          

..............   
for +- 10

we need to calculate probability for ,                                      
190   ≤ X ≤    210                              
X1 =    190   ,    X2 =   210                      
                                      
Z1 =   (X1 - µ )/(σ/√n) = (   190   -   200   ) / (   50   / √   100   ) =   -2.00
Z2 =   (X2 - µ )/(σ/√n) = (   210   -   200   ) / (   50   / √   100   ) =   2.00
                                      
P (   190   < X <    210   ) =    P (    -2.0   < Z <    2.0   )   
                                      
= P ( Z <    2.00   ) - P ( Z <   -2.00   ) =    0.97725   -    0.02275   =    0.9545  
excel formula for probability from z score is =NORMSDIST(Z)                                      
.................

3)

for +-0.03

population proportion ,p=   0.4                      
n=   200                      
                          
std error , SE = √( p(1-p)/n ) =    0.0346                      
                          
we need to compute probability for                           
0.37   < p̂ <   0.43                  
                          
Z1 =( p̂1 - p )/SE= (   0.37   -   0.4   ) /    0.0346   =   -0.866
Z2 =( p̂2 - p )/SE= (   0.43   -   0.4   ) /    0.0346   =   0.866
P(   0.37   < p̂ <   0.43   ) =    P[( p̂1-p )/SE< Z <(p̂2-p)/SE ]    =P(    -0.866   < Z <   0.866   )          
                                                  
= P ( Z <   0.866   ) - P (    -0.866   ) =    0.8068   -   0.193   =   0.6135             

.........

for +- 0.05

we need to compute probability for                           
0.35   < p̂ <   0.45                  
                          
Z1 =( p̂1 - p )/SE= (   0.35   -   0.4   ) /    0.0346   =   -1.443
Z2 =( p̂2 - p )/SE= (   0.45   -   0.4   ) /    0.0346   =   1.443
P(   0.35   < p̂ <   0.45   ) =    P[( p̂1-p )/SE< Z <(p̂2-p)/SE ]    =P(    -1.443   < Z <   1.443   )  
                                          
= P ( Z <   1.443   ) - P (    -1.443   ) =    0.9255   -   0.074   =   0.8511      

.......................

Please revert back in case of any doubt.

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