[tex]\large \bold {SOLUTION}[/tex]
Given values
- Population Mean = 6
- Standard Deviation = 5.3
- Population = 350
- Confidence Level = 99%
Let the value of α = 0.01
To find the length of an interval, we have to find the difference between the upper and lower interval. That is the given formula for both is the same but each mean is added or subtracted between the quotient of the standard deviation and the square root of the population
For the value of the z-score, refer to the z-score table that corresponds to the confidence level.
[tex]\sf{LCI =( \overline{x } + Z_{ \frac{ \alpha }{2} }( \dfrac{\sigma}{ \sqrt{n} } )) - (\overline{x } - Z_{ \frac{ \alpha }{2} }( \dfrac{\sigma}{ \sqrt{n} } ))}[/tex]
[tex]\sf{LCI =( 6 + 2.576( \dfrac{5.3}{ \sqrt{350} } )) - (6 - 2.576( \dfrac{5.3}{ \sqrt{350} } ))}[/tex]
[tex]\sf{LCI =6.73 - 5.27}[/tex]
[tex]\boxed{\green{\sf{LCI\approx1.46}}}[/tex]
❤
