One method for estimating the availability of office space in large cities is to conduct a random sample of offices, and calculate the proportion of offices currently being used. Suppose that real estate agents believe that of all offices are currently occupied, and decide to take a sample to assess their belief. They are considering a sample size of n = 40.
a) Show that this sample size is large enough to justify using the normal approximation to the sampling distribution of p. (2 pts)
b) What is the mean of the sampling distribution of p if the real estate agents are correct? (2pts)
c) What is the standard deviation of the sampling distribution of p if the real estate agents are correct? (4 pts)
d) If the real estate agents are correct, what is the probability that a sample proportion, p, would differ from by as much as 0.05? (8 pts)
a)
b)1
c)standard deviation=
c) 0.07246
a) To determine if the sample size is large enough to justify using the normal approximation, we need to check if the conditions for the normal approximation are satisfied. The conditions are:
1. The sampling method should be simple random sampling - From the problem statement, it is mentioned that the sample is taken using a random sample of offices, which satisfies this condition.
2. The sample size should be large enough - Let's check if the sample size is sufficiently large. The formula for determining the minimum sample size to ensure the normal approximation is given by n(p)(1-p)≥10, where n is the sample size and p is the proportion. Plugging in the values, we get:
n(0.5)(1-0.5)≥10
n(0.25)≥10
n≥10/0.25
n≥40
Since the sample size (n = 40) is larger than or equal to the minimum required sample size (n≥40), we can conclude that the sample size is large enough to justify using the normal approximation.
b) The mean of the sampling distribution of p is equal to the population proportion, which is given as 0.5 (or 50%, since p is the proportion of offices currently occupied).
Mean (μ) = p = 0.5
c) The standard deviation of the sampling distribution of p can be calculated using the formula:
Standard deviation (σ) = √[(p)(1-p)/n]
Plugging in the values, we get:
Standard deviation (σ) = √[(0.5)(1-0.5)/40]
= √[(0.25)/40]
= √[0.00625]
≈ 0.079
Therefore, the standard deviation of the sampling distribution of p is approximately 0.079.
d) To calculate the probability that a sample proportion, p, would differ from the population proportion, 0.5, by as much as 0.05, we need to find the probability that |p - 0.5| ≤ 0.05.
Using the normal distribution, we can convert the values to z-scores and find the area under the curve between the z-scores.
z1 = (0.5 - 0.05 - 0.5) / 0.079 ≈ -0.633
z2 = (0.5 + 0.05 - 0.5) / 0.079 ≈ 0.633
Using a standard normal distribution table or calculator, we can find the respective probabilities corresponding to these z-scores:
P(-0.633 ≤ Z ≤ 0.633) = P(Z ≤ 0.633) - P(Z ≤ -0.633)
Using the standard normal distribution table, the probability for Z ≤ 0.633 is approximately 0.736 and the probability for Z ≤ -0.633 is approximately 0.264.
So, P(-0.633 ≤ Z ≤ 0.633) = 0.736 - 0.264 ≈ 0.472
Therefore, if the real estate agents are correct, the probability that a sample proportion, p, would differ from 0.5 by as much as 0.05 is approximately 0.472.
a) To determine whether the sample size is large enough to justify using the normal approximation to the sampling distribution of p, we need to check if the conditions for using the normal distribution are satisfied. The conditions are:
1. Random Sample: The real estate agents are planning to take a random sample, which is a good practice to ensure representative data.
2. Independence: The sample observations must be independent. This means that the occupancy status of one office should not affect the status of another. It can be assumed that the offices are independent of each other.
3. Success/Failure Condition: We need to check if both np and n(1-p) are greater than or equal to 10, where n is the sample size and p is the estimated proportion of offices being occupied.
In this case, we have n = 40 and p = 0.8 (estimated proportion of offices being occupied).
np = 40 * 0.8 = 32
n(1-p) = 40 * (1 - 0.8) = 8
Since both np and n(1-p) are greater than or equal to 10, the success/failure condition is satisfied. Therefore, the sample size of n = 40 is large enough to justify using the normal approximation to the sampling distribution of p.
b) The mean of the sampling distribution of p can be calculated using the formula:
mean (μ) = p
In this case, if the real estate agents' belief is correct and p = 0.8, then the mean of the sampling distribution of p would be equal to 0.8.
c) The standard deviation of the sampling distribution of p can be calculated using the formula:
standard deviation (σ) = sqrt((p * (1 - p)) / n)
In this case, if the real estate agents' belief is correct and p = 0.8, and the sample size is n = 40, then the standard deviation of the sampling distribution of p would be:
standard deviation (σ) = sqrt((0.8 * (1 - 0.8)) / 40) = sqrt(0.16 / 40) = sqrt(0.004) = 0.0632 (rounded to four decimal places)
d) To calculate the probability that a sample proportion, p, would differ from by as much as 0.05, we need to find the area under the normal curve between two points.
First, we need to standardize the difference of 0.05 using the formula:
z = (x - μ) / σ
In this case, x = 0.05 (difference in proportions), μ = 0.8 (mean), σ = 0.0632 (standard deviation).
z = (0.05 - 0.8) / 0.0632 = -12.17 (rounded to two decimal places)
Next, we need to find the area to the left of z = -12.17 using a z-table or a statistical software. Since the z-value is extremely large, the probability is effectively 0 (very close to 0).
Therefore, if the real estate agents are correct and p = 0.8, the probability that a sample proportion, p, would differ from by as much as 0.05 is essentially 0.