Complete Question
The complete question is shown on the first uploaded image
Answer:
The confidence interval is [tex]64.86<\mu<67[/tex]
Step-by-step explanation:
From the question we are given the following data
The following heights are
66, 65, 67, 62, 62, 65, 61, 70, 66, 66, 71, 63, 69, 65, 71, 66, 66, 69, 68, 62, 65, 67, 65, 71, 65, 70, 62, 62, 63, 64, 67, 67
The sample size is n =32
The confidence level is [tex]k = 95[/tex]% = 0.95
The mean is evaluated as
[tex]\= x = 66+ 65+ 67+ 62+ 62+ 65+ 61+ 70+ 66+ 66+ 71+63+ 69+ 65+ 71+ 66+ 66+ 69+ 68+ 62+ 65+ 67,+\\65+ 71+ 65+ 70+ 62+ 62+ 63+ 64+ 67+ 67 / 32[/tex]
=> [tex]\= x = \frac{2108}{32}[/tex]
=> [tex]\= x = 65.875[/tex]
The standard deviation is evaluated as
[tex]\sigma = \sqrt{ v}[/tex]
Now
[tex]v = ( 66-65.875 )^2+(65-65.875)^2+( 67-65.875)^2+ (62-65.875)^2+ (62-65.875)^2+ (65-65.875)^2+( 61-65.875)^2+ (70-65.875)^2+ (66-65.875)^2+ (66-65.875)^2+ (71+63-65.875)^2+ (69-65.875)^2+ (65-65.875)^2+ (71-65.875)^2+( 66-65.875)^2+ (66-65.875)^2+ (69-65.875)^2+ (68-65.875)^2+ (62-65.875)^2+ (65-65.875)^2+ (67-65.875)^2,+\\(65-65.875)^2+ (71-65.875)^2+ (65-65.875)^2+ (70-65.875)^2+( 62-65.875)^2+( 62-65.875)^2+ (63-65.875)^2+ (64-65.875)^2+ (67-65.875)^2+ (67-65.875)^2 / 32[/tex]
=>[tex]v= 8.567329[/tex]
=> [tex]\sigma = \sqrt{8.567329}[/tex]
=> [tex]\sigma = 2.927[/tex]
The level of significance is evaluated as
[tex]\alpha = 1 - 0.95[/tex]
[tex]\alpha = 0.05[/tex]
The degree of freedom is evaluated as
[tex]Df = n- 1 \equiv Df = 32 -1 = 31[/tex]
The critical values for the level of significance is obtained from the z -table as
[tex]t_c = t_{\alpha/2 } , Df = t _{0.05/2}, 31 =\pm 1.96[/tex]
The confidence interval is evaluated as
[tex]\mu = \= x \pm t_c * \frac{\sigma }{\sqrt{n} }[/tex]
substituting values
[tex]\mu =65.875 \pm 1.96* \frac{2.927}{\sqrt{32} }[/tex]
[tex]\mu =65.875 \pm 1.01415[/tex]
=> [tex]64.86<\mu<67[/tex]
