Answer:
The 90% confidence level is [tex]19.15< L < 20.85[/tex]
Step-by-step explanation:
From the question we are told that
The sample size is [tex]n = 64[/tex]
The mean age is [tex]\= x = 20 \ years[/tex]
The standard deviation is [tex]\sigma = 4 \ years[/tex]
Generally the degree of freedom for this data set is mathematically represented as
[tex]df = n - 1[/tex]
substituting values
[tex]df = 64 - 1[/tex]
[tex]df = 63[/tex]
Given that the level of confidence is 90% the significance level is mathematically evaluated as
[tex]\alpha = 100 - 90[/tex]
[tex]\alpha =[/tex]10 %
[tex]\alpha = 0.10[/tex]
Now [tex]\frac{\alpha }{2} = \frac{0.10}{2} = 0.05[/tex]
Since we are considering a on tail experiment
The critical value for half of this significance level at the calculated degree of freedom is obtained from the critical value table as
[tex]t_{df, \frac{ \alpha}{2} } = t_{63, 0.05 } = 1.669[/tex]
The margin for error is mathematically represented as
[tex]MOE = t_{df , \frac{\alpha }{2} } * \frac{\sigma}{\sqrt{n} }[/tex]
substituting values
[tex]MOE = 1.699 * \frac{4 }{\sqrt{64} }[/tex]
[tex]MOE = 0.85[/tex]
he 90% confidence interval for the true average age of all students in the university is evaluated as follows
[tex]\= x - MOE < L < \= x + E[/tex]
substituting values
[tex]20 - 0. 85 < L < 20 + 0.85[/tex]
[tex]19.15< L < 20.85[/tex]
