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I've got a problem where I'm supposed to find the probability of a point estimate but cannot see how my answer is differing from the given one. The problem is:

Unknown to an experimenter, the probability of a prototype etching procedure producing a defective part is p = 0.24. The experimenter examines 100 randomly selected parts and finds out whether or not each one is defective. What is the probability that the experimenter’s point estimate of p is within 0.05 of the true value?

My approach here is that these are independent binomial counts so they are normally distributed using the mean and variance from the binomial distribution:

$$N\left(p, \frac{p(1-p)}{n}\right)$$

From here we have the following as the expected distribution of 100 samples:

$$N\left(0.24, \frac{0.24(1-0.24)}{100}\right) = N\left(0.24, 0.001824\right)$$

This means for 0.05 around the mean, we have a range of 0.19 to 0.29, the probability for which ought to be calculated by:

$$\phi\left(\frac{0.29-0.24}{\sqrt{0.001824}}\right) - \phi\left(\frac{0.19-0.24}{\sqrt{0.001824}}\right)$$

This gives an answer of 0.75829, but this is not the case, and the answer given is 0.8022.

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    $\begingroup$ The usual notation is to use $\phi$ for the standard normal density function and $\Phi$ for the standard normal cumulative distribution function, intended here. $\endgroup$ – BruceET Jul 3 '15 at 22:00
  • $\begingroup$ Thanks...yeah, I'm relatively new to this notation. $\endgroup$ – Topher Jul 4 '15 at 0:12
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Your idea is substantially correct. I suppose the text intended you to consider the binomial distribution of the number $X$ of successes that would give $\hat p = X/100$ in the desired range. Then using the approximate normal distribution $Norm(\mu = 24, \sigma = \sqrt{100(.24)(1-.24)}),$ with the continuity correction, gives a slightly more precise approximation.

This computation would give $\Phi(\frac{29.5 - 24}{4.271})-\Phi(\frac{18.5 - 24}{4.271}) = 0.80218$. Perhaps a little different if you round as necessary to use normal tables, but clearly in agreement with the answer provided. In R statistical software the computation is:

 diff(pnorm(c(18.5, 29.5), 24, sqrt(18.24)))
 ## 0.8021863

However, if you're going to use statistical software, you might as well avoid the normal approximation and get the exact binomial value. For $X \sim Binom(100, .24)$, you seek $P(19 \leq X \leq 20) = P(18 < X \leq 20) = 0.80297.$ In R we have:

 diff(pbinom(c(18,29),100,.24))   # difference of CDFs
 ## 0.8029728
 sum(dbinom(19:29, 100, .24))     # alternatively, sum of PDFs
 ## 0.8029728

One is lucky to get two-place accuracy with a normal approximation. When feasible, the continuity correction helps. In practice, it is seldom important to have more than two-place accuracy. Nevertheless, in practice nowadays, very few applied statisticians would use a normal approximation when an exact computation is as easy as it is here.

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  • $\begingroup$ Ah yes! I forgot to compensate for continuity....ugh! Thanks! And yeah, it clearly makes sense to use the binomial distribution. For some reason I get caught up in the methods defined in the book, when I've got stats software right here (scipy in python) that can accurately compute this. I guess that just comes with some level of experience. ;) $\endgroup$ – Topher Jul 4 '15 at 0:11
  • $\begingroup$ It's good to know about the normal approx. to the binomial for theoretical and some practical reasons, so you did not waste your time with that. As you learn the topics you will see many connections between distributions. In applied work, and sometimes to get clues about theory, exact computations and even simulations can be useful. With time you will find the right balance of methods for your interests. $\endgroup$ – BruceET Jul 4 '15 at 2:22

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