Question

In the third quarter of 2020, 14.3 million people aged 15 and over were employed in health occupations in the EU, representing 7% of employed people. 13% of EU healthcare workers were doctors. Suppose that a sample of 450 healthcare workers across the EU is observed. 55 of these healthcare workers were doctors. a. Compute the sample proportion of doctors in the observed sample. The sample proportion is (Round to two decimal places.) b. If the population proportion of doctors among healthcare workers is 0.13 , what is the probability to observe a sample of 450 healthcare workers with more than 55 doctors in it? Use the distribution of the sample proportion to answer. Probability is (Round to two decimal places.) c. Compute the probability that the sample proportion of doctors in a sample of 450 healthcare workers is within 0.02 from the population proportion of 0.13. Probability is (Round to two decimal places.)

Answer 1

a. The sample proportion of doctors in the observed sample is 0.12. b. The probability of observing a sample with more than 55 doctors is 68.11%. c. The probability that the sample proportion is within 0.02 from the population proportion is 0.11%.

Step-by-step explanation:

a. The sample proportion of doctors in the observed sample is calculated by dividing the number of doctors in the sample by the total sample size. In this case, there were 55 doctors out of 450 healthcare workers in the sample. Therefore, the sample proportion of doctors is 55/450 = 0.1222, which rounded to two decimal places is 0.12.

b. To calculate the probability of observing a sample with more than 55 doctors, we can use the normal distribution approximation. We can calculate the z-score for 55 doctors using the formula: z = (x - μ) / σ, where x is the observed value, μ is the mean, and σ is the standard deviation. In this case, the mean is n * p = 450 * 0.13 = 58.5, and the standard deviation is sqrt(n * p * (1 - p)) = sqrt(450 * 0.13 * (1 - 0.13)) = 7.46. Plugging these values into the formula, we get z = (55 - 58.5) / 7.46 = -0.47. We can then use a standard normal distribution table or a calculator to find the probability of observing a z-score less than -0.47, which is approximately 0.3189. Therefore, the probability of observing a sample with more than 55 doctors is 1 - 0.3189 = 0.6811, or 68.11% rounded to two decimal places.

c. To calculate the probability that the sample proportion is within 0.02 from the population proportion, we can use the normal distribution approximation. We need to find the z-scores for the lower and upper bounds of the interval. The lower bound is p - 0.02 = 0.13 - 0.02 = 0.11, and the upper bound is p + 0.02 = 0.13 + 0.02 = 0.15. We can then calculate the corresponding z-scores using the formula: z = (p - μ) / σ, where p is the population proportion, μ is the mean, and σ is the standard deviation. In this case, the mean is n * p = 450 * 0.13 = 58.5, and the standard deviation is sqrt(n * p * (1 - p)) = sqrt(450 * 0.13 * (1 - 0.13)) = 7.46. Plugging these values into the formula, we get z_lower = (0.11 - 58.5) / 7.46 = -7.47 and z_upper = (0.15 - 58.5) / 7.46 = -7.20. We can then use a standard normal distribution table or a calculator to find the probabilities corresponding to these z-scores. The probability between these two z-scores is approximately 0.0011. Therefore, the probability that the sample proportion is within 0.02 from the population proportion is approximately 0.0011, or 0.11% rounded to two decimal places.