Answer :
Solution :
a). Let x denotes ACT scores.
ACT scores = [tex]$\{ 22,28,20,21,28,23,26 \} ; n = 7$[/tex]
Mean, [tex]$(\overline x)=\frac{\sum x_i}{n}$[/tex]
[tex]$=\frac{168}{7}=24$[/tex]
Sample variance, [tex]$S^2=\frac{1}{n-1}\left(\sum x_i^2-n\overline x^2 \right)$[/tex]
[tex]$=\frac{1}{6}(4098-7 \times 24^2)$[/tex]
[tex]$=\frac{66}{6}$[/tex]
= 11
b). To test whether or not variance of ACT scores of population (say [tex]$\sigma^2$[/tex]) of the UTC students is significantly more than 8.
Consider the hypothesis :
[tex]$H_0: \sigma^2 \leq8$[/tex] vs [tex]$H_a: \sigma^2 >8$[/tex]
It is a right tailed test and α = 0.05
We have
[tex]$x^2_{n-1} = \frac{(n-1)s^2}{\sigma^2}$[/tex]
So test statics is
[tex]$x^2_7=\frac{(7-1)11}{8}$[/tex]
[tex]$=\frac{6 \times 11}{8}$[/tex]
= 8.25
Since our [tex]\text{test statistics is less than the critical value }[/tex]and it falls in a acceptation region, hence we fail to reject [tex]$H_0$[/tex] and conclude that variance is not greater than 8 significantly.