Question

In a study of monthly salary distribution of residents in Paris conducted in year 2015, it was found that the salaries had an average of €2200 (EURO) and a standard deviation of €550. Assume that the salaries were normally distributed.

Question 1: Consider sampling with sample size 64 on the above population. Compute the mean of the sampling distribution of the mean (?̅).

Question 2: Compute the standard deviation of the sampling distribution of the mean in Question 1 above.

Question 3: A random sample of 64 salaries (sample 1) was selected from the above population. What is the probability that the average of the selected salaries is above €2330?

Question 4: Would the calculation you performed in Question 3 still be valid if the monthly salaries were NOT normally distributed? Why? In another study conducted in the same year (2015), the average monthly salary of residents in Bordeaux was found to be about €2353. And the standard deviation of the monthly salaries was €420. A random sample of 81 salaries (sample 2) was selected from this population. Set 1 = Paris (2015); 2 = Bordeaux (2015)

Question 5: Compute the mean of ?̅ 1 − ?̅ 2.

Question 6: Compute the standard deviation of ?̅ 1 − ?̅ 2.

Question 7: What is the probability that the average of the salaries in the sample 1 is less than the average of the salaries in sample 2? In 2017, a study on the salary distribution of Paris residents was conducted. Assume that the salaries were normally distributed. A random sample of 10 salaries was selected and the data are listed below: 3200 3500 3000 2100 2950 2050 2440 3100 3500 2500

Question 8: Assume that the standard deviation of the salaries was still the same as in 2015. Estimate the average salary (in 2017) with 95% confidence.

Question 9: The assumption made in Question 8 was certainly unrealistic. Estimate the average salary (in 2017) with 95% confidence again assuming that the standard deviation had changed from 2015.

Question 10: Estimate the variance of monthly salaries of Paris residents (in 2017) based on the sample provided above at a 95% confidence level.

Question 11: How would you interpret the result in Question 10 above? A similar study was conducted on salary distribution of Paris residents in 2019. The research team aimed to estimate the average salary. They chose the 98% confidence and assumed that the population standard deviation was the same as in 2015. Assume again that those salaries were normally distributed.

Question 12: If they would like the (margin of) error to be no more than €60, how large a sample would they need to select?

Answer 1

Solution :

Let X represents the monthly salary distribution of residents in Paris for the year 2015.

Given that,

Mean (μ) = €2200

Standard deviation (σ) = €550

If X is a normally distributed population with mean μ and standard deviation σ, then sampling distribution of sample mean (x̄) follows normal distribution with mean μ and standard deviation σ/√n. (Where, n is sample size).

i.e.

1) We have sample size (n) = 64 and μ = €2200

Mean of the sampling distribution of mean (x̄) will be €2200.

2) We have, sample size (n) = 64 and  σ = €550

Standard deviation of the sampling distribution of sample mean (x̄) will be given as follows :

Hence, Sstandard deviation of the sampling distribution of sample mean (x̄) is €68.75.

3) We have to obtain P(x̄ > €2330).

Where, x̄ is the average salary of the randomly selected 64 salaries.

We have, μ = €2200,  σ = €550 and n = 64

And also

We know that if x̄ ~ N(μ, σ​​​​​​2/n) then

Using "pnorm" function of R we get, P(Z > 1.8909) = 0.0293

Hence, the probability that that the average of the selected salaries is above €2330 is 0.0293.

4) According to central limit theorem, if we have a population (not normally distributed) with mean μ and standard deviation σ, and if we take samples of sufficiently large sizes (say n = at least 30) then, sampling distribution of sample mean (x̄) follows approximately normal distribution with mean μ and standard deviation σ/√n. (Where, n is sample size).

We have, n = 64, which is sufficiently large, therefore sampling distribution of our sample mean will be approximately normally distributed with mean μ and standard deviation σ/√n. This is same as we have done in the question 3. Therefore, the calculation that I performed in the question 3 will still be valid.