Question #187435

Assuming the standard deviation is unchanged and that the usage is normally distributed, provide an expression for calculating a 99% confidence interval for the mean usage in the March quarter of 2006. (Exactly one option must be correct)



1
Expert's answer
2021-05-07T10:28:32-0400

Since σσ is assumed known,


we use the interval xˉ±zσn,x̄ ± z \dfrac{σ}{ n},


 where σ=81,n=30σ = 81, n = 30 and where z is chosen to ensure that P(Zz)=0.99.P(|Z|≤ z) = 0.99.


From the normal tables, P(Z2.575)=0.99P(|Z|≤ 2.575) = 0.99 (because P(Z<2.575)=0.995)P(Z < 2.575) = 0.995) and so we use z=2.575.z = 2.575.


99% confidence interval expression =xˉ±2.575×σn= \bar{x}\pm 2.575\times \dfrac{\sigma}{n}


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