A surveyor is measuring the height of a cliff known to be about 1000 feet. He as

Question

A surveyor is measuring the height of a cliff known to be about 1000 feet. He assumes his instrument is properly calibrated and that his measurement errors are independent, with mean mu = 0 and variance sigma^2 = 10. He plans to take n measurements and form the average. Estimate, using (a) Chebyshev's inequality and (b) the normal approximation, how large n should be if he wants to be 95 percent sure that his average falls within 1 foot of the true value. Now estimate, using (c) Chebyshev's inequality and (d) the normal approximation, how small sigma^2 must be if he wants to take only 10 measurements with the same resulting confidence.

Explanation / Answer

Solution:

From the given information we have Mean = 0, Variance 2 = 10
Given level of significance is = 0.05
Here we want to find the values of sample size n such that
P(-0.01 < x- < 0.01) 0.95
=> P(|x- | > 0.01) 1- 0.95

By Chebyshev's inequality we have,
P(| x- | > t) 2/nt2
That implies

P(| x- | > 0.01) 0.05
=> 10/n*0.012 = 0.05
=> n = 2,000,000

b) Here we want to find the value of sample size n using normal approximation.
n = (Z/2/e)2
From the normal area table values Z0.05/2 = 1.96
n = (1.96*3.1623/0.01)
= 384165.33
c) Here we need to find the value for when the sample size n = 10
By Chebyshev's inequality we have
P(| x- | > t) 2/nt2
=> 2/10*0.012 = 0.05
2 = 0.00005
= 0.007
using normal approximation :
n = (Z/2/E)2
here Z0.05/2 = 1.96
10 = (1.96 * /0.01)2
=> = sqrt(10*0.012/1.962)
=> = 0.0161

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.