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Sagot :
Answer:
A. 0.0272
B. 0.0278
C. No
Step-by-step explanation:
a. Calculation for the margin of error and the interval estimate of the number of eligible people who had a driver’s license in 1995u
First step is to calculate the Standard error using this formula
Standard error(SE)= √(pcap * (1 - pcap)/n)
Where,
pcap represent sample proportion = 0.639
n represent sample size = 1200
Let plug in the formula
Standard error (SE)= √(0.639 * (1 - 0.639)/1200)
Standard error(SE)= 0.0139
Second step is to find Zc using this formula
Zc = Z(α/2)
Where,
α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025
Zc = Z(α/2) = 1.96
Now let calculate the Margin of Error using this formula
Margin of Error= zc * SE
Let plug in the formula
Margin of Error = 1.96 * 0.0139
Margin of Error = 0.0272
Therefore the Margin of Error is 0.0272
b. Calculation for the margin of error and the interval estimate of the number of eligible people who had a driver’s license in 2016
First step is to calculate the Standard error using this formula
Standard error(SE)= √(pcap * (1 - pcap)/n)
Where,
pcap represent sample proportion = 0.417
n represent sample size = 1200
Let plug in the formula
Standard error (SE)= √(0.417 * (1 - 0.417)/1200)
Standard error(SE)= 0.0142
Second step is to find Zc using this formula
Zc = Z(α/2)
Where,
α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025
Zc = Z(α/2) = 1.96
Now let calculate the Margin of Error using this formula
Margin of Error= zc * SE
Let plug in the formula
Margin of Error = 1.96 * 0.0142
Margin of Error = 0.0278
Therefore the Margin of Error is 0.0278
c. Based on the above calculation for both (a) and (b) the MARGIN OF ERROR are NOT the same reason been that (a) Margin of Error is 0.0272 while (b) Margin of Error is 0.0278 which simply shows that the proportion in parts (a) and are different.
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