We take a simple random sample of 100 people and check their ages. Out of the 100, 15 are 65. What is the probability that the proportion of people over 65 in the entire population is between 13% and 17%? What error range can we guarantee with 90% certainty?
Use the normal approximation to the binomial distribution.
mean = np
sd = √npq (where q = 1-p)
Therefore:
mean = 100 * .15 = 15
sd = √npq = √(100 * .15 * .85) = 3.57
Use z-scores:
z = (13 - 15)/3.57 = -0.56
z = (17 - 15)/3.57 = 0.56
Use a z-table to find the probability between the two z-scores.
For the second part, use a 90% confidence interval formula.
I hope this will help get you started.
To determine the probability that the proportion of people over 65 in the entire population is between 13% and 17%, we need to use statistical inference. Specifically, we can use the concept of confidence intervals.
To calculate the probability, we'll use the binomial distribution since we have a fixed sample size (100) and a binary outcome (over 65 or not over 65). The sample proportion can be calculated as the number of successes (15) divided by the sample size (100), which is 0.15 or 15%.
To find the probability that the proportion falls between 13% and 17%, we need to calculate the cumulative probability of the binomial distribution within that range. We can use software or statistical tables to obtain the exact probability. Let's assume that the probability is P, so we have:
P(13% ≤ proportion ≤ 17%) = P(proportion ≤ 17%) - P(proportion < 13%)
Next, we need to determine the standard error of the proportion, which quantifies the uncertainty in our estimate. The formula for the standard error (SE) of a proportion can be calculated as:
SE = sqrt((proportion * (1 - proportion)) / sample size)
In this case, it would be SE = sqrt((0.15 * 0.85) / 100) ≈ 0.0387.
Using the standard error, we can calculate the margin of error (ME) for a 90% confidence level. The formula for the margin of error is:
ME = z * SE
Where z represents the critical value for a given confidence level. For a 90% confidence level, the z-value is approximately 1.645.
Applying the formula, we have:
ME = 1.645 * 0.0387 ≈ 0.0636
Thus, we can guarantee a 90% confidence interval by adding and subtracting the margin of error (ME) to the sample proportion. Therefore, the error range with 90% certainty is:
15% ± 0.0636
Finally, to find the probability of the proportion falling within this interval, we need to calculate the cumulative probability of the binomial distribution within this range, as mentioned earlier.