Confidence intervals. A student is suspicious of the formula of confidence inter
ID: 3176034 • Letter: C
Question
Confidence intervals. A student is suspicious of the formula of confidence intervals her sloppy professor wrote down on a napkin over lunch. She decided to conduct an experiment to check on the correctness of the formula. She generated random samples from computer and used the professor's formula to calculate a 95% confidence interval from each sample. Then she checks for each sample if the true parameter falls within the confidence interval. After checking 97 samples, the confidence interval contains the true parameter 92 times. "Ha, " she said, "now the confidence interval has to contain the true parameters in all next three times to be correct." When it turns out that in the next 3 samples, none of the confidence intervals contain the true parameter, she gleefully told her classmates that the professor made another mistake. Do you agree with her? Why and why not? What is the probability that a 95% confidence interval contains the true parameter in exactly 95 out of 100 random samples?Explanation / Answer
The 95% confidence interval interprets that 95% of the intervals obtained from random samples generated will contain the true population parameter. Therefore, out of 97 samples, if 92 times the interval contains the true population parameter, then the percentage is:92/97*100=94.84~95. Therefore, the students is not right.
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There are n=100 independent, random trials, with probability of success, p=0.9485. The probability of success is constant throughout the trials. The specific number of success in n trials is denote dby r. Use, Binom(100, 0.9485) distribution to compute the required probability using the follwoing formula.
P(X,r)=nCr(p)^r(1-p)^n-r [nCr=n!/r!(n-r)!]
P(X=95)=100C95(0.9485)^95(1-0.9485)^5=0.1796 (ans)