The principal of the school thinks that the average IQ of students at is at least 110.

This question involves a test of hypothesis,  therefore, the first step is to decide the null and the alternative hypothesis to be used. As usual, the null hypothesis will always contain the quality sign. 

Hypotheses

H0: mu = 100

Ha: mu < 110>

using an alpha of 0.05,

t = (x_bar - mu)/(s/sqrt(n))

t = (108 - 110)/(10/sqrt(20)) = -0.894

The critical value for this test can be obtained in excel using the t.inv() Function,  =T.INV(0.05,19) = -1.729

The criteria is to reject th null hypothesis if the test test statistic (t - test) which follows a t distribution is less than the critical value (-1.729)

Alternatively, the p -value for this test can be calculated in excel using the = T-DIST() function, =T.DIST(-0.894,19,TRUE) = 0.191, 

And the null hypothesis is rejected in the p value is less than, In this case, the p value is greater than 0.05, therefore, the null hypothesis is not rejected. There is not sufficient evidence to reject the professor's claim.