Question

The average first salary (that is the salary received at their first job after finishing college), in the young population of one country in Europe follows a normal distribution with mean 1100 euros per month and standard deviation 200 euros. We want to verify the hypothesis that the first salary of young people in that country who have a masters' degree from a foreign university, is not different than the population average. We look at a random sample of 300 young people in that European country. The average first salary in the sample is 1120 euros per month. Compute the test statistic Z which you need to test the hypothesis.

Please explain on how to use the table if needed.

Options available:

a. z=1,645

b. z=0.04

c. z=0.025

d. z=0.975

Answer 1

Given:

Population mean, =1100; Population standard deviation, =200; Sample mean, =1120; Sample size, n =300

The test statistic is:

Z = = =1.732​​​​​​

The critical value of of Z at 5% significance level ( =0.05) for two-tailed test(to know if sample mean is significantly different from population mean) is: Z =1.96​​​​​​

[Z_1- =Z_1-0.025 =Z_0.975 =1.96: from cumulative Z table, the corresponding Z value for 0.975 area is: 1.96. Since, it is a two tailed test, we have Z= -1.96 at the left tail and Z= +1.96 at the right tail].

If we want to test if sample mean is significantly greater than the population mean (right tailed test), then the critical value of Z at 5% significance level is: Z_crit =1.645​​​​​​

[Z =Z_1-0.05 =Z_0.95 =1.645: from cumulative Z table, the corresponding Z value for 0.95 area is +1.645 (at right tail)].

Thus, option a. Z=1.645 is correct regarding the critical Z-value for right tailed test.

At 0.975, the corresponding Z =1.9 +0.06 =1.96

At 0.95 (we have 0.9495 and 0.9505 in the table) the corresponding Z =1.6 +0.045 =1.645