A population proportion has been estimated at 0.32 calculate the following with a sample size of 160.

Question

A find the probability of getting a sample proportion at most 0.32
B find the probability of getting a sample proportion at least 0.36?

Answer 1

A) To find the probability of getting a sample proportion at most 0.32, we can use the standard normal distribution and the formula for the z-score:

z = (p̂ - p) / sqrt(p * (1-p) / n)

where p̂ is the sample proportion, p is the population proportion, and n is the sample size. In this case, p = 0.32, p̂ = 0.32, and n = 160. Plugging in these values, we get:

z = (0.32 - 0.32) / sqrt(0.32 * 0.68 / 160) = 0

Since the z-score is 0, the probability of getting a sample proportion at most 0.32 is simply the area to the left of 0 on the standard normal distribution, which is 0.5.

B) To find the probability of getting a sample proportion at least 0.36, we again use the standard normal distribution and the formula for the z-score:

z = (p̂ - p) / sqrt(p * (1-p) / n)

In this case, p = 0.32, n = 160, and we want to find the probability of getting a sample proportion greater than or equal to 0.36. To find the corresponding z-score, we first find the standard error of the sample proportion:

SE = sqrt(p * (1-p) / n) = sqrt(0.32 * 0.68 / 160) = 0.035

Then, we can calculate the z-score:

z = (0.36 - 0.32) / 0.035 = 1.14

Using a standard normal table or calculator, we can find the area to the right of 1.14, which is 0.1271. However, since we want the probability of getting a sample proportion at least 0.36, we need to add the area to the left of -1.14 (since the standard normal distribution is symmetric). The area to the left of -1.14 is also 0.1271, so we can add these two areas to get the final probability:

0.1271 + 0.1271 = 0.2542

Therefore, the probability of getting a sample proportion at least 0.36 is 0.2542.