2 A random sample of size n 289 from a N mu sigma2 distrib

2. A random sample of size n = 289 from a N ( mu, sigma^2) distribution gives the sample mean bar x = 789 and the sample standard deviation s = 34. If the number of degrees of freedom ( d.f.) is large, the values of ta/2 can be approximated by za/2. Compare the values of z to t for z 0.05 = 1.645 to the value oft 0.05 (n - 1 d.f) z 0.025 = 1.96 to the value oft 0.025 (n - 1 d.f.). Also construct a confidence interval for 90% and 95% on mu

Solution

For 90% confidence interval:

The degree of freedom =n-1= 289-1=288

t(0.05, df=288) = 1.65 (from student t table)

So the lower bound is

xbar - t*s/vn = 789-1.65*34/sqrt(289) =785.7

So the upper bound is

xbar+ t*s/vn = 789+1.65*34/sqrt(289) =792.3

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For 95% confidence interval:

The degree of freedom =n-1= 289-1=288

t(0.025, df=288) = 1.97 (from student t table)

So the lower bound is

xbar - t*s/vn = 789-1.97*34/sqrt(289) =785.06

So the upper bound is

xbar+ t*s/vn = 789+1.97*34/sqrt(289) =792.94