A simple random sample of size n is drawn from a normally distributed population, and the mean of the sample is x overbar, while the standard deviation is s. what is the 99% confidence interval for the population mean? use the table below to help you answer the question. confidence level 90% 95% 99% z*-score 1.645 1.96 2.58 x overbar plus-or-minus startfraction 0.90 times s over startroot n endroot endfraction x overbar plus-or-minus startfraction 0.99 times s over startroot n endroot endfraction x overbar plus-or-minus startfraction 1.645 times s over startroot n endroot endfraction x overbar plus-or-minus startfraction 2.58 times s over startroot n endroot endfraction

A confidence interval is a set of calculations for an unrecognized parameter that is specified as a range with a lower and upper bound.

If we take an n-sample from a normal population and wish to estimate the population mean, the confidence interval for the mean is:

[tex]\bar x[/tex]±[tex]Z_\frac{\alpha }{2} \times \frac{s}{\sqrt{n} }[/tex]

The level of significance is found by;

[tex]\rm \alpha =1- \ condifidence\ level \\\\\ \rm \alpha =1- 0.99 \\\\\ \rm \alpha = 0.001[/tex]

The value of z is

[tex]\rm Z_{\frac{\alpha}{2}} = \rm Z_{\frac{0.01}{2}} = Z_{0.005}=2.58[/tex]

Hence the 99% confidence interval for the population mean will be x± [tex]\frac{2.58 s}{\sqrt{n} }[/tex].

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