Stats help please

Question

^Assume that a sample is used to estimate a population portion P find a 99% confidence interval for a sample of size 286 with 39% successes. Enter your answer as a try linear inequality using decimals not percent accurate to three decimal places.

Answer 1

Let's solve this by determining what we are looking for, what information we are working with, and, given these criteria, the appropriate formula to achieve our goal.

1) First, we can ascertain that we will estimate a population parameter from a sampling distribution because that is what confidence intervals tell us - the estimated true population parameter and a margin of error surrounding our estimate.

2) Next, we can ask a few questions that may guide our choice of formula to apply and reassure us that we did step one correctly. Importantly, we can ask, "What data are we working with?" To answer this question, we can break this down further into:

(a) "What type of data are we working with to find our centrality measure?," - discrete data -;

(b) "Is this a hypothesis test?," - no it is not -, and;

(c) "Do we know the standard deviation of the population?," - no we do not.

Our answers to these questions tell us:

(a) that we will be working with a proportion and the estimated mean proportion rather than just an estimated mean;

(b) that we will be estimating a mean proportion rather than a difference in means and there should be no "p < ..." for what the scope of what this question has asked of us, and ;

(c) that we will be estimating the sampling distribution of the population mean using a sample statistic, which, for proportions, doesn't affect the z distrubution we will be using; however, it is a good question to be in the habit of asking

3) We then can find our 99% confidence interval using our estimated spread of the sampling distribution of the proportion mean √((p'•q')/n), where p' is the sample, or estimated, proportion and q' = 1-p'. Then we multiply that by the standard deviations in the standard normal of z* away from the mean that are at .005 and .995 because we are splitting our 100%-99%=1% into the "low end'" of values below 0.5 % (.005) and the "high end" of values above 99.5% (.995).

Our z values/quantiles are then ± 2.575829

Applying the above:

p' = .39

q' = .61

n = 286

CI = p' ± z_α/2 • √((p'•q')/n)

.39 - 2.575829 • √((.39 * .61)/286) < P < .39 - 2.575829 • √((.39 * .61)/286)

.3157009 < P < .4642901

NOTE

1. For proportions, there needs to be a sufficient number of observations of successes and failures. Books will range from at least 5 to at least 15 (never less than 5 per group). Another calculation can be used.
2. If the interval [p - 3√((p'•q')/n), p + 3√((p'•q')/n)] lies between 0 and 1 there are considered to be enough observation in the sample for an accurate population estimate.