Answer:
The 99% confidence interval is [tex]97.94 < \mu < 98.26[/tex]
Step-by-step explanation:
From the question we are told that
The sample size is n = 110
The sample mean is [tex]\= x = 98.1 \ F[/tex]
The standard deviation is [tex]\sigma = 0.64 \ F[/tex]
Given that the confidence level is 99% the level of significance i mathematically evaluated as
[tex]\alpha = 100 - 99[/tex]
[tex]\alpha = 1\%[/tex]
[tex]\alpha = 0.01[/tex]
Next we obtain the critical value of [tex]\frac{\alpha }{2}[/tex] from the normal distribution, the values is
[tex]Z_{\frac{\alpha }{2} } = Z_{\frac{0.01 }{2} } = 2.58[/tex]
Generally the margin of error is mathematically represented as
[tex]E = Z_{\frac{\alpha }{2} } * \frac{\sigma}{\sqrt{n} }[/tex]
substituting values
[tex]E = 2.58 * \frac{ 0.64}{\sqrt{110} }[/tex]
[tex]E = 0.1574[/tex]
Generally the 99% confidence interval is mathematically represented as
[tex]\= x - E < \mu < \= x + E[/tex]
substituting values
[tex]98.1 - 0.1574 < \mu < 98.1 + 0.1574[/tex]
[tex]97.94 < \mu < 98.26[/tex]
