Answer:
The test statistic  t = 1.219 < 2.262 at 5% level of significance
we accept significance of  level that the population mean is less than 20.
Step-by-step explanation:
Given ages are 22, 17, 27, 20, 23, 19, 24, 18, 19, and 24 
The random sample (n) = 10
Null hypothesis (H0): μ = 20
Alternative hypothesis(H1) : μ < 20 (left tailed test)
we will use statistic 't' distribution with small sample 10 < 30 
 
mean (χ) = sum of observations divided by no of observations
mean(x) = ∑x / n = 
x         x - mean           (x-mean)^2
22      22-21.3  =  0.7       0.49
17       17-21.3    = -4.3      18.49
27      27-21.3 = 5.7         32.49
20      20-21.3 =-1.3          1.69
23      23-21.3 = 1.7          2.89
19       19-21.3 = -2.3        5.29
24       24-21.3 = 2.7       7.29
18        18-21.3 = -3.3      10.89
19        19-21.3 = -2.3      5.29
24       24-21.3 = -=2.7    7.29
                                    
                                    
                                   S = 3.198 
The test statistic (t) = 
                             t = 1.219
The degrees of freedom = n-1 = 10-1 =9
Tabulated value of t for 9 degrees of freedom at 5% level of significance
= 2.262
since calculated t < tabulated t we accept the null hypothesis
we accept significance of  level that the population mean is less than 20.