Question

Consider the next \({\rm{1000}}\) 95% CIs for m that a statistical consultant will obtain for various clients. Suppose the data sets on which the intervals are based are selected independently of one another. How many of these\({\rm{1000}}\)intervals do you expect to capture the corresponding value of m? What is the probability that between \({\rm{940 and 960}}\)Do these intervals contain the corresponding value of m? (Hint: Let Y = the number among the \({\rm{1000}}\) intervals that contain m. What kind of random variable is Y?)

Short answer

\(P(940 \le Y \le 960) = 0.853\)

Step-by-step solution

Step 1: 

Random variable Y is a binomial random variable with parameters \({\rm{1000}}\) and \({\rm{p = 95\% = 0}}{\rm{.95}}\) Proposition: For a binomial random variable X with parameters n, p, and q=\(1\)-p, the following is true

\(\begin{array}{l}{\rm{E(X) = npn}}\\{\rm{V(X) = np(1 - p) = npqn}}\\{{\rm{\sigma }}_{\rm{X}}}{\rm{ = }}\sqrt {{\rm{npq}}} \end{array}\)

Therefore, the expected number among the 1000 to capture mu is

\({\rm{E(Y) = n \times p = 1000 \times 0}}{\rm{.95 = 950}}\)

In order to calculate the probability, the standard deviation is required to obtain the normalization of a binomial random variable. The standard deviation is

\({{\rm{\sigma }}_{\rm{Y}}}{\rm{ = }}\sqrt {{\rm{npq}}} {\rm{ = }}\sqrt {{\rm{1000 \times 0}}{\rm{.95 \times 0}}{\rm{.05}}} {\rm{ = 6}}{\rm{.892}}\)

To find the probability

Finally, the probability is

\(\begin{array}{l}P(940 \le Y \le 960) = P\left( {\frac{{940 - 950}}{{6.892}} \le \frac{{Y - E(Y)}}{{{\sigma _Y}}} \le \frac{{960 - 950}}{{6.892}}} \right)\\ = P( - 1.45 \le Z \le 1.45)\\ = P(Z \le 1.45) - P(Z \le - 1.45)\end{array}\)

from the normal probability table in the appendix. The probability can also be computed with software.

Hence the kind of random variable y is

\(P(940 \le Y \le 960) = 0.853\)