Complete Question
The complete question is shown on the first uploaded image
Answer:
The 99% confidence interval is [tex]2.4309 < \mu < 4.9328[/tex]
Step-by-step explanation:
From the question we are told that
The data is 3, 2,5, 3, 2, 6, 5,4.5, 3, 3, and 4
The sample size is n = 11
Generally the sample mean is mathematically represented as
[tex]\= x = \frac{\sum x_i}{n}[/tex]
=> [tex]\= x = \frac{3+ 2+5+ \cdots +4 }{11}[/tex]
=> [tex]\= x = 3.6818 [/tex]
Generally the sample standard deviation is mathematically represented as
[tex]\sigma = \sqrt{\frac{\sum [ x_i - \= x ] }{n} }[/tex]
=> [tex]\sigma = \sqrt{\frac{[ 3 - 3.6818 ]^2 +[ 2 - 3.6818 ]^2 + \cdots + [ 4 - 3.6818 ]^2 }{11} }[/tex]
=> [tex]\sigma = 1.3091[/tex]
Given that the confidence interval is 99% then the level of significance is mathematically represented as
[tex]\alpha= (100 - 99) \%[/tex]
=> [tex]\alpha= 0.01[/tex]
Given that the sample size is small we will making use of t distribution table
Generally from the t distribution table the critical value of at a degree of freedom of is
[tex]t_{\frac{\alpha }{2} , 10 } =3.16927267 [/tex]
Generally the margin of error is mathematically represented as
[tex]E = 3.16927267 * \frac{1.3091 }{\sqrt{11} }[/tex]
[tex]E = 1.251[/tex]
Generally 99% confidence interval is mathematically represented as
[tex]3.6818 -1.251 < p < 3.6818 + 1.251 [/tex]
=> [tex]2.4309 < \mu < 4.9328[/tex]
