A population proportion has been estimated at 0.32 calculate the following with a sample size of 160.
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?
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.
To calculate the probabilities, we need to use the sampling distribution of the sample proportion, which follows an approximately normal distribution when the sample size is large enough. The mean of the sampling distribution is the true population proportion (0.32 in this case), and the standard deviation is given by the formula:
Standard deviation (σ) = sqrt((p * q) / n)
where p is the population proportion, q is 1-p, and n is the sample size.
In this case, p = 0.32, q = 1 - 0.32 = 0.68, and n = 160. Let's calculate the required probabilities step by step:
A. Probability of getting a sample proportion at most 0.32:
To find this probability, we need to find the area under the normal distribution curve to the left of 0.32. We can use the Z-score formula to standardize the value:
Z = (x - μ) / σ
where x is the sample proportion, μ is the population proportion, and σ is the standard deviation.
In this case, x = 0.32, μ = 0.32, and σ = sqrt((0.32 * 0.68) / 160) ≈ 0.032.
Z = (0.32 - 0.32) / 0.032 = 0
Since the Z-score is 0, the probability of getting a sample proportion at most 0.32 is equal to the area to the left of Z = 0, which is 0.5.
Therefore, the probability of getting a sample proportion at most 0.32 is 0.5.
B. Probability of getting a sample proportion at least 0.36:
To find this probability, we need to find the area under the normal distribution curve to the right of 0.36. Again, we can use the Z-score formula.
In this case, x = 0.36, μ = 0.32, and σ ≈ 0.032 (as calculated in part A).
Z = (0.36 - 0.32) / 0.032 ≈ 1
The area to the right of Z = 1 can be found using a Z-table or a statistical software. Using a standard normal distribution table, the area to the right of Z = 1 is approximately 0.1587.
Therefore, the probability of getting a sample proportion at least 0.36 is approximately 0.1587.