Suppose that a statistical consultants\' job is to compute 95% confidence interv

Question

Suppose that a statistical consultants' job is to compute 95% confidence intervals for mu for different clients. The context in which the clients collect data varies (as does the sample size and population mean) but the confidence level (95%) is always the same. (a) Last year the consultant computed 500 such confidence intervals. Let X be the number of times his confidence interval actually contained the true population parameter mu. Find the distribution of X. (b) Use the Central Limit Theorem to find an approximation for the probability that between 470 and 495 intervals contain the population mean mu. Use a continuity correction if appropriate.

Explanation / Answer

Part-a

From the definition of 95% confidence interval we know that if we construct a large number of confidence intervals then approximately 95% of them contains the true value of population mean.

So, X follows binomial distribution with n=500 and p=0.95

As n is large so using central limit theorem, X follows normal distribution with

Mean=np=500*0.95 =475

Standard deviation = sqrt(500*0.95*(1-0.95))=4.87

Part-b

Using continuity correction we have

P(470<X<495)=P(469.5<X<495.5)

=0.8706 using excel function =NORMDIST(495.5,475,4.87,TRUE)-NORMDIST(469.5,475,4.87,TRUE)

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