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?

Answer 1

Soln

Given,

Population Mean = 2200

Population Std Dev = 550

1)

Mean of Sampling Distribution = 2200

2)

Sample Size (n) = 64

Sample Std Dev = Population Std Dev / n1/2 = 68.75

3)

P(X>2330) = 1 – P(X<2330) = 1 – P(z<(2330-2200)/ 68.75)

= 1 – P(z<1.89) = 1 - 0.9706 = 0.0294 or 2.94% (Using the z table)

4)

Yes, the calculations we did in Question 3 will still be valid as our sample size is large enough (greater than 30) and we can assume the normality assumptions.

5)

Group 1 (Paris)

Population Mean (X1) = 2200

Population Std Dev (S1) = 550

Sample Size (n1) = 64

Group 2 (Bordeaux)

Population Mean (X2) = 2353

Population Std Dev (S2) = 420

Sample Size (n2) = 81

X1-X2 = 2200 – 2353 = -153

6)

Assuming, the population std deviation is not same.

SPooled = (S12/n1 + S22/n2 )1/2 = 83.09

7)

Null and Alternate Hypothesis

H0: µ1 = µ2

Ha: µ1 < µ2

Test Statistic

Assuming, the population std deviation is not same.

Z = (X1 ­­­– X2 ­– (µ1 - µ2))/ (s12/n1 + s22/n2 )1/2 = -153/83.09 = -1.84

P(Z<-1.84) = 0.0329 or 3.29%

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