Question

Question 8 1 pts Suppose these following 10 numbers are random drawn from a Normallu. o ? ) distribution with known g 2 = 0.5 (4.7, 5.5, 4.4, 3.3, 4.6, 5.3, 5.2, 4.8, 5.7, 5.3) p is believed to have a non-informative prior (say Normal with a really high variance). What is the chance that u will be between 4.442 and 5.318? That's a invalid question. mu can not be bounded by two numbers 95% 90%

Answer 1

Solution:

We are given

Lower limit = 4.442

Upper limit = 5.318

Margin of error = (Upper limit – lower limit) / 2

Margin of error = (5.318 – 4.442)/2

Margin of error = 0.877/2

Margin of error = 0.4385

Margin of error = Z*σ/sqrt(n)

We have σ = sqrt(0.5) = 0.707107

Sample size = n = 10

Margin of error = Z*σ/sqrt(n)

0.4385 = Z*0.707107/sqrt(10)

Z = 0.4385*sqrt(10)/0.707107

Z = 1.961031

So, confidence level = 0.95 or 95% approximately

(by using z-table)

Answer: 95%