Answer:
The 93% confidence interval is [tex] 26.52 < \mu < 33.48 [/tex]
Step-by-step explanation:
From the question we are told that
The sample size is n = 41
The mean is [tex]\= x = 30[/tex]
The standard deviation is [tex]\sigma = 12.3[/tex]
From the question we are told the confidence level is 93% , hence the level of significance is
[tex]\alpha = (100 - 93 ) \%[/tex]
=> [tex]\alpha = 0.07[/tex]
Generally from the normal distribution table the critical value of [tex]\frac{\alpha }{2}[/tex] is
[tex]Z_{\frac{\alpha }{2} } = 1.81[/tex]
Generally the margin of error is mathematically represented as
[tex]E = Z_{\frac{\alpha }{2} } * \frac{\sigma }{\sqrt{n} }[/tex]
=> [tex]E = 1.812 * \frac{12.3}{\sqrt{41} }[/tex]
=> [tex]E = 3.48 [/tex]
Generally 93% confidence interval is mathematically represented as
[tex]\= x -E < \mu < \=x +E[/tex]
=> [tex] 30 -3.48 < \mu < 30 + 3.48 [/tex]
=> [tex] 26.52 < \mu < 33.48 [/tex]
