Question

Suppose I want to construct a confidence interval on the mean run time of a bit of optimization code that I have written. Suppose we know that the run time is approximately normal with variance 225 minutes. I run the code 20 times and find the sample mean of the run times is 325 minutes. If I were to construct a 99% confidence interval of the form

L <= µ <= U

What would be the value of L?

Answer 1

Answer:

a) The 99% of Control limits are

316.36 < µ< 333.64

b) Lower limit  :  = 316.36

Step-by-step explanation:

Explanation:-

Step(i)

Given the sample size 'n' = 20

Given the sample mean 'x⁻' = 325minutes

Suppose we know that the run time is approximately normal with variance 225 minutes

Population variance 'σ²' = 225 minutes

Standard deviation of the population 'σ' = √225 = 15

Level of significance ∝ = 0.99

Step(ii):-

99% of confidence intervals of the mean is determined by

$(x^{-} - z_{\alpha } \frac{S.D}{\sqrt{n} } , x^{-} + z_{\alpha } \frac{S.D}{\sqrt{n} })$

The z-score of 99% of confidence intervals =2.576

$(325 - 2.576\frac{15}{\sqrt{20} } , 325+ 2.576 \frac{15}{\sqrt{20} })$

on calculation, we get

(325 - 8.640 , 325 + 8.640)

99% of Control limits are

316.36 < µ< 333.64

Lower limit  : 325 - 8.640 = 316.36

upper limit  : 325 + 8.640 = 333.64

Conclusion:-

a) The 99% of Control limits are

316.36 < µ< 333.64

b) Lower limit  :  = 316.36