Answer:
The 90% confidence interval is [tex] 2.47<\mu < 2.59 [/tex]
Step-by-step explanation:
From the question we are told that
The sample size is n = 6
The sample mean is [tex]\= x = 2.53 \ mm[/tex]
The standard deviation is [tex]\sigma = 0.090\ mm[/tex]
Given that the confidence level is 90% then the level of significance is
[tex]\alpha = (100 - 90)\%[/tex]
=> [tex]\alpha = 0.10 [/tex]
Generally from the normal distribution table the critical value of [tex]\frac{\alpha }{2}[/tex] is
[tex]Z_{\frac{\alpha }{2} } = 1.645 [/tex]
Generally the margin of error is mathematically represented as
[tex]E =1.645 * \frac{0.090 }{\sqrt{6} }[/tex]
=> [tex]E = 0.060 [/tex]
Generally 90% confidence interval is mathematically represented as
[tex]\= x -E < \mu < \=x +E[/tex]
[tex]2.53 -0.060 <\mu < 2.53 + 0.060[/tex]
=> [tex] 2.47<\mu < 2.59 [/tex]
