Answer :
Answer:
The 95% confidence interval is [tex] 2.354 < 3.526 [/tex]
Step-by-step explanation:
From the question we are told that
The sample size is n = 5
The sample data is 3.32 2.53 3.45 2.38 3.01
Gnerally the sample mean is mathematically represented as
[tex]\= x = \frac{\sum x_i}{n}[/tex]
=> [tex]\= x = \frac{3.32+ 2.53 +3.45+ 2.38+ 3.01 }{5}[/tex]
=> [tex]\= x = 2.94 [/tex]
Generally the standard deviation is mathematically represented as
[tex]s = \sqrt{ \frac{ \sum (x_i - \= x )^2}{n-1} }[/tex]
=> [tex]s = \sqrt{ \frac{ ( 3.32 - 2.94 )^2 + ( 2.53 - 2.94 )^2+ \cdots + ( 3.01 - 2.94 )^2}{5-1} }[/tex]
=> [tex]s = 0.472 [/tex]
Generally the degree of freedom is mathematically represented as
[tex]df = n-1[/tex]
=> [tex]df = 5-1[/tex]
=> [tex]df = 4 [/tex]
From the question we are told the confidence level is 95% , hence the level of significance is
[tex]\alpha = (100 - 95 ) \%[/tex]
=> [tex]\alpha = 0.05[/tex]
Generally from the t distribution table the critical value of [tex]\frac{\alpha }{2}[/tex] at a degree of freedom of [tex]df = 4 [/tex] is
[tex]t_{\frac{\alpha }{2} ,df } = 2.77[/tex]
Generally the margin of error is mathematically represented as
[tex]E =t_{\frac{\alpha }{2} ,df } * \frac{\sigma }{\sqrt{n} }[/tex]
[tex]E = 2.77 * \frac{0.472 }{\sqrt{5} }[/tex]
[tex]E = 0.586 [/tex]
Generally 95% confidence interval is mathematically represented as
[tex]\= x -E < \mu < \=x +E[/tex]
[tex]2.94-0.586 < 2.94 + 0.586[/tex]
[tex] 2.354 < 3.526 [/tex]