Question

11. For a random sample of size 13 from a normal distribution with mean u, you are given the following regarding the observations: (ti – 1)2 = 77.8 The width of the 100% confidence interval for u is 2.7005. Let tay be the critical value of a t random variable with v degrees of freedom. The following table lists values of tay for specific combinations of a and v: v = 12 v=13 a=0.1 1.356 1.350 a= 0.07 1.580 1.572 a= 0.04 1.912 1.899 Find the value of k.

Answer 1

Solution:

We are given

Width of confidence interval = 2.7005

Half width = margin of error = E = 2.7005/2 = 1.35025

Sample size = n = 13

Variance = ∑(X – Xbar)^2/(n – 1) = 77.8/(13 – 1) = 77.8/12 = 6.483333

Standard deviation = S = sqrt(variance) = sqrt(6.483333) = 2.546239

Degrees of freedom = n – 1 = 13 – 1 = 12

Margin of error = t*S/sqrt(n)

Margin of error = 1.35025

1.35025 = t*2.546239/sqrt(13)

1.35025 = t*0.7062

t = 1.35025/0.7062

t = 1.911994

t = 1.912

df = ν = 12

So, α = 0.04

(by using given table)

k = 1 – α = 1 – 0.04 = 0.96

k = 0.96