Answer:
a
[tex]\= x = 18.5[/tex] , [tex]\sigma = 5.15[/tex]
b
[tex]15.505 < \mu < 21.495[/tex]
c
[tex]14.93 < \mu < 22.069[/tex]
Step-by-step explanation:
From the question we are are told that
The sample data is 21, 14, 13, 24, 17, 22, 25, 12
The sample size is n = 8
Generally the ample mean is evaluated as
[tex]\= x = \frac{\sum x }{n}[/tex]
[tex]\= x = \frac{ 21 + 14 + 13 + 24 + 17 + 22+ 25 + 12 }{8}[/tex]
[tex]\= x = 18.5[/tex]
Generally the standard deviation is mathematically evaluated as
[tex]\sigma = \sqrt{\frac{\sum (x- \=x )^2}{n}}[/tex]
[tex]\sigma = \sqrt{\frac{\sum ((21 - 18.5)^2 + (14-18.5)^2+ (13-18.5)^2+ (24-18.5)^2+ (17-18.5)^2+ (22-18.5)^2+ (25-18.5)^2+ (12 -18.5)^2 )}{8}}[/tex]
[tex]\sigma = 5.15[/tex]
considering part b
Given that the confidence level is 90% then the significance level is evaluated as
[tex]\alpha = 100-90[/tex]
[tex]\alpha = 10\%[/tex]
[tex]\alpha = 0.10[/tex]
Next we obtain the critical value of [tex]\frac{ \alpha }{2}[/tex] from the normal distribution table the value is
[tex]Z_{\frac{ \alpha }{2} } = 1.645[/tex]
The margin of error is mathematically represented as
[tex]E = Z_{\frac{ \alpha }{2} } * \frac{\sigma }{\sqrt{n} }[/tex]
=> [tex]E =1.645 * \frac{5.15 }{\sqrt{8} }[/tex]
=> [tex]E = 2.995[/tex]
The 90% confidence interval is evaluated as
[tex]\= x - E < \mu < \= x + E[/tex]
substituting values
[tex]18.5 - 2.995 < \mu < 18.5 + 2.995[/tex]
[tex]15.505 < \mu < 21.495[/tex]
considering part c
Given that the confidence level is 95% then the significance level is evaluated as
[tex]\alpha = 100-95[/tex]
[tex]\alpha = 5\%[/tex]
[tex]\alpha = 0.05[/tex]
Next we obtain the critical value of [tex]\frac{ \alpha }{2}[/tex] from the normal distribution table the value is
[tex]Z_{\frac{ \alpha }{2} } = 1.96[/tex]
The margin of error is mathematically represented as
[tex]E = Z_{\frac{ \alpha }{2} } * \frac{\sigma }{\sqrt{n} }[/tex]
=> [tex]E =1.96 * \frac{5.15 }{\sqrt{8} }[/tex]
=> [tex]E = 3.569[/tex]
The 95% confidence interval is evaluated as
[tex]\= x - E < \mu < \= x + E[/tex]
substituting values
[tex]18.5 - 3.569 < \mu < 18.5 + 3.569[/tex]
[tex]14.93 < \mu < 22.069[/tex]
