Answer:
The 95% CI is [tex]2.108 < \mu < 2.892[/tex]
Step-by-step explanation:
From the question we are told that
The population mean [tex]\mu = 2.5[/tex]
The standard deviation is [tex]\sigma = 0.8[/tex]
Given that the confidence level is 95% then the level of confidence is mathematically 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 values is [tex]Z_{\frac{\alpha }{2} } = 1.96[/tex]
Generally the margin of error is mathematically evaluated as
[tex]E = Z_{\frac{\alpha }{2} } * \frac{\sigma}{\sqrt{n} }[/tex]
here we would assume that the sample size is n = 16 since the person that posted the question did not include the sample size
So
[tex]E = 1.96* \frac{0.8}{\sqrt{16} }[/tex]
[tex]E = 0.392[/tex]
The 95% CI is mathematically represented as
[tex]\= x -E < \mu < \= x +E[/tex]
substituting values
[tex]2.5 - 0.392 < \mu < 2.5 + 0.392[/tex]
substituting values
[tex]2.108 < \mu < 2.892[/tex]
