Answer: The standard deviation of test scores in the class is not less than 14.1
Step-by-step explanation:
Let's suppose that the test scores follow a normal distribution. Besides, we have:
a) Standard deviation [tex]s=12.3[/tex]
b) Significance level [tex]\alpha =.05[/tex]
c) n=27
Using a) we can deduce that sample variance [tex]s^{2} = s*s = 151.29[/tex].
Since we want to prove if the population variance is less than [tex]14.1^{2}[/tex]:
[tex]H_{0}[/tex] (Null hypotesis) : [tex]\sigma^{2} =(14.1)^{2}[/tex]
[tex]H_{1}[/tex] (Alternative hypotesis): [tex]\sigma^{2} \leq (14.1)^{2}[/tex]
For test this kind of hypotesis (variance in one population) the correct test statistic is [tex]((n-1)s^{2})/\sigma^{2}[/tex], which under [tex]H_{0}[/tex] have Chi-square distribution with n-1 degrees of freedom.
Calculating the test statistic ([tex]\sigma^{2}[/tex] is the value in [tex]H_{0}[/tex] ) :
[tex]\frac{(27-1)*(151.29)}{(14.1)^2} = 19.79[/tex]
For this hypotesis (left one tailed test) the p-value is [tex]P(M<19.79)[/tex] where M follow a Chi-square distribution with n-1=26 degrees of freedom.You can check in a chi-square table that p-value=0.1986
If [tex]pvalue>\alpha[/tex] then there is no evidence to statistically reject [tex]H_{0}[/tex] . Therefore, the standard deviation of test scores in the class is not less than 14.1 (95% confidence level).
