Answer:
a. Error is 0.14
b. between 1913.68 to 1772.32
Step-by-step explanation:
Requirement a)
The margin of error is a statistic expressing the amount of random sampling error in a survey's results.
Given,
Size of the sample n = 50
We know, At 95% confidence interval,
The margin of error,
= 0.98/√n
= 0.98/√50
= 0.98/7.07
= 0.13859 ~ 0.14
So the margin of error at 95% confidence level is 0.14.
Requirement b)
Given,
Standard deviation σ = 255
Population mean z = 1814
Size of the sample n = 50
We know, At 95% confidence interval,
(z-µ)/(σ/√n) < Z_0.05
= - Z_0.05 < (z-µ)/(σ/√n) < Z_0.05
= - 1.96 * σ/√n < (z-µ)/(σ/√n)* σ/√n < 1.96 * σ/√n [ in two tail Z test the value of Z_0.05 is 1.96 ]
= - 1.96 * 255/√50 < Z-µ < 1.96 * 255/√50
= -70.6828 < Z-µ < 70.6828
= 70.6828 > µ - Z > -70.6828
= 70.6828 + Z > µ > -70.6828 + Z
= 70.6828 + 1814 > µ > -70.6828 + 1814
=1913.68 > µ > 1772.32
So, the sample mean would be between 1913.68 to 1772.32.
