A hotel manager believes that 28% of the hotel rooms are booked. If the manager is correct, what is the probability that the proportion of rooms booked in a sample of 895 rooms would differ from the population proportion by less than 3% ? Round your answer to four decimal places.
The population proportion of rooms booked is 28%, so the population standard deviation can be calculated as:
σ = sqrt((p * (1 - p)) / n) = sqrt((0.28 * (1 - 0.28)) / 895) ≈ 0.0148
To find the probability that the proportion of rooms booked in a sample of 895 rooms would differ from the population proportion by less than 3%, we need to find the area under the normal curve within ±3% (0.03) of the population proportion (0.28).
First, we calculate the z-score for the lower bound:
z_lower = (0.28 - 0.03 - 0.28) / 0.0148 = -2.03
Next, we calculate the z-score for the upper bound:
z_upper = (0.28 + 0.03 - 0.28) / 0.0148 = 2.03
Using a standard normal distribution table or calculator, we can determine the proportion within these bounds.
P(-2.03 < Z < 2.03) ≈ 0.9772
Rounding to four decimal places, the probability that the proportion of rooms booked in a sample of 895 rooms would differ from the population proportion by less than 3% is approximately 0.9772.
To calculate the probability, we will use the formula for calculating the standard error of the sample proportion:
SE = sqrt((p * q) / n)
where p is the population proportion, q is 1 - p, and n is the sample size.
Given:
p = 0.28 (population proportion)
q = 1 - p = 1 - 0.28 = 0.72
n = 895 (sample size)
Calculating the standard error (SE):
SE = sqrt((0.28 * 0.72) / 895)
SE = sqrt(0.2016 / 895)
SE = sqrt(0.000225504)
SE ≈ 0.015021308
Next, we need to find the range within which the proportion can differ from the population proportion by less than 3%:
3% of 0.28 = 0.03 * 0.28 = 0.0084
So, the range is θ = 0.0084.
To find the probability, we will use the normal distribution. Since the sample size is large (n > 30), we can assume a normal distribution for the sample proportion.
Now, we need to find the z-scores corresponding to the lower and upper bounds of the range:
Lower bound z-score: z1 = (0.0084 - 0) / 0.015021308 = 0.5594 (rounded)
Upper bound z-score: z2 = (-0.0084 - 0) / 0.015021308 = -0.5594 (rounded)
Using a standard normal distribution table or a calculator, we can find the corresponding probabilities for these z-scores:
P(Z < 0.5594) = 0.7123 (from the table)
P(Z < -0.5594) = 0.2883 (from the table)
To find the probability that the proportion would differ by less than 3%, we subtract the lower probability from the upper probability:
P(-0.0084 < p - 0.28 < 0.0084) = P(Z < 0.5594) - P(Z < -0.5594)
= 0.7123 - 0.2883
= 0.4240
Therefore, the probability that the proportion of rooms booked in a sample of 895 rooms would differ from the population proportion by less than 3% is approximately 0.4240.
To find the probability that the proportion of rooms booked in a sample of 895 rooms would differ from the population proportion by less than 3%, we can use the sampling distribution of proportions and then calculate the z-score.
Step 1: Calculate the standard deviation of the sampling distribution.
The standard deviation of the sampling distribution, also known as the standard error, can be calculated using the formula:
Standard Error = sqrt((p * (1 - p)) / n)
Where:
- p is the proportion of rooms booked (0.28)
- n is the sample size (895)
Let's calculate it:
Standard Error = sqrt((0.28 * (1 - 0.28)) / 895) = sqrt(0.2022 / 895) = sqrt(0.000226)
Step 2: Calculate the z-score.
The z-score represents the number of standard deviations away from the mean.
z = (Sample Proportion - Population Proportion) / Standard Error
In this case, we want the sample proportion to differ from the population proportion by less than 3%, which means we need to find the z-score for a difference of 0.03 (3% expressed in decimal form):
z = (0.03 - 0.28) / sqrt(0.000226)
Let's calculate it:
z = -0.25 / sqrt(0.000226) = -0.25 / 0.01503 = -16.62
Step 3: Find the probability using the standard normal distribution table.
The standard normal distribution table gives us the probability for a given z-score. In this case, we want to find the probability that the z-score is less than -16.62.
Looking up the z-score in the standard normal distribution table, we find that the probability is essentially 0.
Therefore, the probability that the proportion of rooms booked in a sample of 895 rooms would differ from the population proportion by less than 3% is approximately 0.0000 (rounded to four decimal places).