If X1 X16 are a random sample from a normal population wit

If X_1, ... , X_16 are a random sample from a normal population with mu unknown and sigma^2 = 9, what is the 95% confidence interval for mu? What is the probability that the sample variance S^2 is greater than or equal to 20 when sigma^2 = 9? If S^2 turns out to be 20, compare the probability that you found with 0.01. Can we insist on the assumption that sigma^2 = 9? If not, what is the 95% confidence interval for mu based on the value of S^2?

Solution

95% confidence interval = [ sample mean - ( 1.96 * 3 / sqrt (16) ) , sample mean + ( 1.96 * 3 / sqrt (16) )..

chi-sq= (n-1) S^2 / sigma^2 = 15* 20 / 9 = 33.333......
probability = p[ s^2 >=12] = p [ chisq >= 33.333 ] = 0.004224459...

S^2 =20...
the prob. we found earlier is less than 0.01..

so, at 0.01 level of significance , we can\'t say that sigma^2 = 9 ..

so, 95% confidence interval for m based on S^2 is :

= [ sample mean - ( 2.13145 * sqrt( 20) / sqrt ( 16) ) , sample mean + ( 2.13145 * sqrt( 20) / sqrt ( 16) ) ]

where, 2.13145 = t-score at 95% confidence.

when we were using sigma for standarization, we were using a z-score(normal dist) , now when we used S i.e the sample s.d for standardization , we have to use t-score( because it follows a t-distribution) !