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 $x=140lb{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}{s}^{2}=225.\text{}\text{Find the}\text{}95\mathrm{\%}(z=2)$ confident interval.

liannemdh

Skilled2021-03-06Added 106 answers

Step 1

The$(1-\alpha )100\mathrm{\%}$ confidence interval formula for the population mean when population standard deviation is not known, is defined as follows:

$CI=\stackrel{\u2015}{x}\pm {t}_{\frac{\alpha}{2},n-1}\left(\frac{s}{\sqrt{n}}\right).$

$t}_{\frac{\alpha}{2},n-1$ is the critical value of the t-distribution with degrees of freedom of $n-1$ above which, $100\left(\frac{\alpha}{2}\right)\mathrm{\%}{\textstyle \phantom{\rule{1em}{0ex}}}\text{or}{\textstyle \phantom{\rule{1em}{0ex}}}\frac{\alpha}{2}$ proportion of the observations lie, and below which, $100(1-\alpha +\frac{\alpha}{2}=100(1-\frac{\alpha}{2})\mathrm{\%}{\textstyle \phantom{\rule{1em}{0ex}}}\text{or}{\textstyle \phantom{\rule{1em}{0ex}}}(1-\frac{\alpha}{2})$ proportion of the observation lie, $\stackrel{\u2015}{x}$ is the sample mean, s is the sample standard deiation, and n is the sample size.

Step 2

The sample mean is$\stackrel{\u2015}{x}=140lb$ , and the sample standard deviation is $s=15lb(=\surd 225)$ .

The sample size is$n=400$ . The degrees of freedom is $399(=400\u20131).$

The confidence level is 0.95. Hence, the level of significance is$1\u20130.95=0.05.$

Using Excel formula:$=T.INV.2T(0.05,399)$ , the critical value is, ${t}_{\frac{\alpha}{2}}={t}_{0.025}\approx 1.96.$

Thus,

$CI=\stackrel{\u2015}{x}\pm {t}_{\frac{\alpha}{2}},n-1\left(\frac{s}{\sqrt{n}}\right)$

$=140\pm \left(1.96\right)\left(\frac{15}{\sqrt{400}}\right)$

$=140\pm \left(1.96\right)\left(0.75\right)$

$=140\pm 1.47$

$=(138.53,141.47)$

Thus, the required$95\mathrm{\%}$ confidence interval is (138.53 lb, 141.47 lb).

The

Step 2

The sample mean is

The sample size is

The confidence level is 0.95. Hence, the level of significance is

Using Excel formula:

Thus,

Thus, the required