Question

In: Statistics and Probability

Consider the 1000 95% confidence intervals (CI) for μ that a statistical consultant will obtain for various clients.

Consider the 1000 95% confidence intervals (CI) for μ 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 1000 intervals do you expect to capture the corresponding value of μ? What is the probability that between 950 and 970 of these intervals contain the corresponding value of μ? (Hint: Let Y = the number among the 1000 intervals that contain μ. What kind of random variable is Y?).

Solutions

Expert Solution

Solution

A binomial distribution describes the possible number of times that a particular event will occur in a sequence of observations.

The conditions for the binomial distribution is,

• A trial has only two possible outcomes namely success or failure.

• There is fixed number  of identical trials.

• The trials of the experiment are independent of each other.

The normal approximation to the binomial is applicable when the number of experiments (sample size) is large and the probability of success is close to 0.5.

Conditions for normal approximation to binomial distribution:

oা< bu 01< du

The probability distribution function of Binomial distribution is,

P(X==)=(*) p*(1pyaan

Here,

 The number of trials

 The probability of success

q=l-p= The probability of failure

The formulas for mean and standard deviation using the normal approximation is,

bdu = 2 du = 11

The formula for  score is,

-* =ج

Step: 1

From the information, the values are given as follows:

n=1000 p=0.95

The expected value to capture the corresponding value of is,

E(X)=1000x0.95 = 950


Explanation:

Out of 100, the expected value (950) to capture the corresponding value of 


Hint:

Check the conditions for normal to binomial approximation is valid for the given problem.

Step: 2

From the information, the values are given as follows:

Conditions for normal approximation to binomial distribution:

np 10 1000 x 0.95 2 10 950 > 10 nq 210 1000x0.05 2 10 50 > 10


Explanation:

Based on the results, observe that both the conditions are satisfied. Hence, use the normal approximation to binomial distribution.


Hint:

Use  to determine the probability that between 950 and 970 of these intervals contain the corresponding value of 

Step: 3

The value of mean is,

v=np = 1000x0.95 = 950

The value of standard deviation is,

o= npg = V1000x0.95 x 0.05 ( 9=1-p) = 47.5 = 6.8920

The probability that between 950 and 970 of these intervals contain the corresponding value of  is,

Pros P(950 < X <970)=P689208 Y - 070-p949.5-950 X- 970.5-950 (Using continuity correction) 6 .8920) = P(Z <-0.073)- P(Z <2.97


Explanation:

About 52.74% of chance that between 950 and 970 of these intervals contain the corresponding value of 

 


The expected value is 950.

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