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.
Originally Posted by randomer86
has a t-distribution with n-1 degrees of freedom. The 90% confidence region
for t with 16 degrees of freedom is:
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.