Question

A random sample of `n=100` is drawn from a Binomial distribution, yielding an average of 0.25. What is the mean and standard deviation of `hatp`? What is the probability that `p` is between 0.18 and 0.46?

Answer 1

The mean of `hatp` is 0.25; the standard deviation (standard error) is 0.0433.

`"P"(0.18< p < 0.46)=0.9471.`

Explanation:

The question should state that the random sample comes from a Bernoulli distribution with `p=0.25.` (If it were Binomial, we would need to know the `n` for that distribution, which is not the `n=100` given.)

b)

If we have 100 `X` variables that are each `"Bernoulli"(p),` then `hatp=barX` is approximately normal with mean `p` and standard deviation `sqrt((p(1-p))/100).`

In this example, that means `hatp` is approximately normal with

`mu=0.25`

and

`sigma = sqrt((0.25xx0.75)/100)=sqrt3/40~~0.0433.`

As an example, when we draw a card from a standard deck, 25% of our draws will be spades, and 75% will be other suits. So if we draw a card 100 times, we should expect `25% xx 100 = 25` spades, and `75` non-spades. Makes sense, right?

In other words, we should expect the proportion in our sample to be fairly close to the proportion in the population: `hatp ~~ p.` The question is: how close?

The answer is: the larger our sample, the closer our estimate. Think about it: if we draw only one card, then `hatp` is `"Bernoulli"(p),` just like each `X`. Thus, for a single draw, `"s.e."(hatp)=sigma=sqrt(p(1-p)).`

But if we could possibly draw a card 1,000,000 times, our value for `hatp` would be almost guaranteed to match `p`, which means `hatp` would have almost no standard error. This is why the standard error formula for `hatp` is

`"s.e."(hatp)=sigma/sqrtn`

In other words: as we draw more cards (`n->oo`), the expected difference between our `hatp` and the real `p` will shrink (`"s.e."(hatp)->0`).

c)

Since `hatp` is approximately `"N"(0.25,0.0433)`, we have

`"P"(0.18 < p < 0.46)`

`="P"((0.18-mu)/sigma < (p-mu)/sigma < (0.46-mu)/sigma)`

`="P"((0.18-0.25)/0.0433 < Z < (0.46-0.25)/0.0433)`

`="P"(–1.617 < Z < 4.850)`

`="P"(Z<4.850)" "-" ""P"(Z<= –1.617)`

`=Phi(4.850)-Phi(–1.617)`

`~~0.9999994-0.05293914`

`~~0.9471`.