 UkusakazaL

2021-08-02

A study a local high school tried to determine the mean height of females in the US. A study surveyed a random sample of 125 females and found a mean height of 64.5 inches with a standard deviation of 5 inches. Determine a $95\mathrm{%}$ confidence interval for the mean.
$\begin{array}{|cc|}\hline \text{Confidence Interval}& z\\ 80\mathrm{%}& 1.282\\ 85\mathrm{%}& 1.440\\ 90\mathrm{%}& 1.645\\ 95\mathrm{%}& 1.960\\ 99\mathrm{%}& 2.576\\ 99.5\mathrm{%}& 2.807\\ 99.9\mathrm{%}& 3.291\\ \hline\end{array}$ ruigE

Step 1
Solution:
Given information:
$n=125$ Sample size
$x=64.5$ inches Sample mean
$s=5$ inches sample standard deviation
$\alpha =0.05$ Level of significance
Step 2
The 95% confidence interval for the mean is
$\stackrel{―}{x}±{t}_{\frac{\alpha }{2},n-1}×\frac{s}{\sqrt{n}}$
n is large we used ${z}_{\frac{\alpha }{2}}$ instead of ${t}_{\frac{\alpha }{2},n-1}$
At $\alpha =0.05$
${z}_{\frac{\alpha }{2}}={z}_{0.05}=1.96$ From Z table
$\left(64.5±1.96×\frac{5}{\sqrt{125}}\right)$
$\left(64.5±1.96×\frac{5}{11.180339}\right)$
$\left(64.5±1.96×0.4472135\right)$
$\left(64.5±0.8765384\right)$
$\left(64.5-0.8765384,64.5+0.8765384\right)$
$\left(63.62346,65.376538\right)$
$\left(63.62,65.38\right)$
The $95\mathrm{%}$ confidence interval for the mean is ( 63.62, 65.38)

Do you have a similar question?