abondantQ

2021-03-05

A random sample of $n=400$ students is selected from a population of $N=4000$ students to estimate the average weight of the students. The sample mean and sample variance are found to be confident interval.

liannemdh

Step 1
The $\left(1-\alpha \right)100\mathrm{%}$ confidence interval formula for the population mean when population standard deviation is not known, is defined as follows:
$CI=\stackrel{―}{x}±{t}_{\frac{\alpha }{2},n-1}\left(\frac{s}{\sqrt{n}}\right).$
${t}_{\frac{\alpha }{2},n-1}$ is the critical value of the t-distribution with degrees of freedom of $n-1$ above which, $100\left(\frac{\alpha }{2}\right)\mathrm{%}\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}\frac{\alpha }{2}$ proportion of the observations lie, and below which, $100\left(1-\alpha +\frac{\alpha }{2}=100\left(1-\frac{\alpha }{2}\right)\mathrm{%}\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}\left(1-\frac{\alpha }{2}\right)$ proportion of the observation lie, $\stackrel{―}{x}$ is the sample mean, s is the sample standard deiation, and n is the sample size.
Step 2
The sample mean is $\stackrel{―}{x}=140lb$, and the sample standard deviation is $s=15lb\left(=\surd 225\right)$.
The sample size is $n=400$. The degrees of freedom is $399\left(=400–1\right).$
The confidence level is 0.95. Hence, the level of significance is $1–0.95=0.05.$
Using Excel formula: $=T.INV.2T\left(0.05,399\right)$, the critical value is, ${t}_{\frac{\alpha }{2}}={t}_{0.025}\approx 1.96.$
Thus,
$CI=\stackrel{―}{x}±{t}_{\frac{\alpha }{2}},n-1\left(\frac{s}{\sqrt{n}}\right)$
$=140±\left(1.96\right)\left(\frac{15}{\sqrt{400}}\right)$
$=140±\left(1.96\right)\left(0.75\right)$
$=140±1.47$
$=\left(138.53,141.47\right)$
Thus, the required $95\mathrm{%}$ confidence interval is (138.53 lb, 141.47 lb).

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