Question

A researcher in Institute of Labour Market Information has claimed that current salary for young graduates is around RM3000. A survey from 200 graduates with one-year experience shows an average salary of RM2900 with standard deviation of RM350. Test the claim of the researcher whether the current salary of young graduates is at least RM3000.

Answer 1

Given:

Population mean (µ) = RM 3000.

Sample size(n) = 200

Sample mean = xbar = RM 2900.

Sample standard deviation = S = RM350

The claim statement is, "the current salary of young graduates is at least RM 3000".

That is µ >= 3000.

Hence, the claim statement goes under the null hypothesis.

Hence, the Null hypothesis is, H0: µ >= 3000

And the alternative hypothesis is, H1: µ < 3000

Here, we have the sample standard deviation. So, we need to use the T-test for testing.

Now, let's find the test statistic.

That is, test statistic T = -4.0406

Degrees of freedom = n - 1 = 200 - 1 = 199.

Now, let's find the P-value.

For finding the P-value, we can use a technology like excel/Ti84 calculator or T table.

The following excel command is used to find the P-value.

= T.DIST(Test statistic, Degrees of freedom, 1)
= T.DIST(-4.0406, 199 , 1)

You will get, P-value = 0.000038

Following is the decision rule for making the decision.

If, P-value > α, then we fail to reject the null hypothesis.

If, P-value <= α, then we reject the null hypothesis.

Here, we are not given the level of significance (alpha).

When alpha is not given then, we generally use alpha =0.05

And P-value = 0.000038

That is P-value < α.

Hence, we reject the null hypothesis.

That is, there is not sufficient evidence to support the claim that the current salary of young graduates is at least RM 3000.