Answer:
The 95% confidence interval is [tex]84.83< \mu < 90.37[/tex]
Step-by-step explanation:
From the question we are told that
The sample size is [tex]n = 113[/tex]
The sample mean is [tex]\= x = 87.6[/tex]
The standard deviation is [tex]\sigma = 15[/tex]
Given that the confidence level is 95% then the level of significance is mathematically represented 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]
Generally the margin of error is mathematically evaluated as
[tex]E = Z_{\frac{\alpha }{2} } * \frac{ \sigma}{ \sqrt{n} }[/tex]
=> [tex]E = 1.96 * \frac{ 15}{ \sqrt{113} }[/tex]
=> [tex]E = 2.77[/tex]
The 95% confidence interval is mathematically represented as
[tex]\= x - E < \mu < \= x + E[/tex]
substituting values
[tex]87.6 - 2.77< \mu < 87.6 + 2.77[/tex]
[tex]84.83< \mu < 90.37[/tex]
