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
![t = \frac{χ-μ}{\frac{S.D}{\sqrt{n-1} } }](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B%CF%87-%CE%BC%7D%7B%5Cfrac%7BS.D%7D%7B%5Csqrt%7Bn-1%7D%20%7D%20%7D)
mean (χ) = sum of observations divided by no of observations
mean(x) = ∑x / n = ![\frac{22+17+27+20+23+19+24+18+19+24}{10} = 21.3](https://tex.z-dn.net/?f=%5Cfrac%7B22%2B17%2B27%2B20%2B23%2B19%2B24%2B18%2B19%2B24%7D%7B10%7D%20%3D%2021.3)
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^{2} = \frac{∑(x-mean)^2}{n-1}= \frac{92.1}{10-1}](https://tex.z-dn.net/?f=S%5E%7B2%7D%20%3D%20%5Cfrac%7B%E2%88%91%28x-mean%29%5E2%7D%7Bn-1%7D%3D%20%5Cfrac%7B92.1%7D%7B10-1%7D)
![S^{2} = \frac{92.1}{9} = 10.2](https://tex.z-dn.net/?f=S%5E%7B2%7D%20%3D%20%5Cfrac%7B92.1%7D%7B9%7D%20%3D%2010.2)
S = 3.198
The test statistic (t) = ![t = \frac{21.3-20}{\frac{3.198}{\sqrt{10-1} } }](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B21.3-20%7D%7B%5Cfrac%7B3.198%7D%7B%5Csqrt%7B10-1%7D%20%7D%20%7D)
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.