Answer:
The 99% confidence interval is [tex]71.67 < \mu < 78.33[/tex]
Step-by-step explanation:
From the question we are told that
The sample size is [tex]n = 15[/tex]
The sample mean is [tex]\= x = 75[/tex]
The standard deviation is [tex]s = 5[/tex]
Given that confidence is 99% then the level of significance is mathematically represented as
[tex]\alpha = 100 - 99[/tex]
[tex]\alpha = 1\%[/tex]
[tex]\alpha = 0.01[/tex]
Next we obtain the critical values of [tex]\frac{ \alpha }{2}[/tex] from the normal distribution table
The value is
[tex]Z_{\frac{ \alpha }{2} } = 2.58[/tex]
Generally the margin for error is mathematically represented as
[tex]E = Z_{\frac{ \alpha }{2} } * \frac{ s}{ \sqrt{n} }[/tex]
=> [tex]E = 2.58 * \frac{ 5}{ \sqrt{15} }[/tex]
=> [tex]E = 3.3307[/tex]
The 99% confidence interval is mathematically represented as
[tex]\= x -E < \mu < \= x +E[/tex]
=> [tex]75 - 3.3307 < \mu <75 + 3.3307[/tex]
=> [tex]71.67 < \mu < 78.33[/tex]
