# Question 1

• Feb 14th 2007, 02:34 PM
randomer86
Question 1
Hi guys, i need help.

A 90% confidence interval for μ based on a sample of size 17 from a Normal distribution has upper end-point equal to 525.9. Given that the sample standard deviation is s = 583, calculate the lower end-point of the interval, correct to 1 decimal place.
• Feb 15th 2007, 08:52 AM
CaptainBlack
Quote:

Originally Posted by randomer86
Hi guys, i need help.

A 90% confidence interval for μ based on a sample of size 17 from a Normal distribution has upper end-point equal to 525.9. Given that the sample standard deviation is s = 583, calculate the lower end-point of the interval, correct to 1 decimal place.

Here we are dealing with the t-distribution.

t=(mu-m)/(s/sqrt(n))

has a t-distribution with n-1 degrees of freedom. The 90% confidence region
for t with 16 degrees of freedom is:

(-1.75, 1/75),

or for mu:

(-1.75 s/sqrt(n) + m, 1.75 s/sqrt(n) + m) = (-1.75x583/sqrt(17)+m, 1.75x583/sqrt(17)+m)
...................= (-247.446+m, 247.466+m)

But we are told that the upper end point is 525.9, so:

525.9=247.466 + m,

so m=278.434, and so the lower end of the interval is: 30.968, or rounding
to 1 decimal place: 31.0.

RonL