Answer:
Check the explanation
Step-by-step explanation:
We want to estimate the total weight of grain on the field based on the data on a simple random sample of 5 piles out of 200. The population and sample sizes are N=200 & n=5 respectively.
1) Let Y_1,Y_2,...,Y_{200} be the weight of grain in the 200 piles and y_1,y_2,...,y_{5} be the weights of grain in the pile from the simple random sample.
We know, the sample mean is an unbiased estimator of the population mean. Therefore,
![\widehat{\mu}=\overline{y}=\frac{1}{5}\sum_{i=1}^{5}y_i=\frac{1}{5}(3.3+4.1+4.7+5.9+4.5)=4.5](https://tex.z-dn.net/?f=%5Cwidehat%7B%5Cmu%7D%3D%5Coverline%7By%7D%3D%5Cfrac%7B1%7D%7B5%7D%5Csum_%7Bi%3D1%7D%5E%7B5%7Dy_i%3D%5Cfrac%7B1%7D%7B5%7D%283.3%2B4.1%2B4.7%2B5.9%2B4.5%29%3D4.5)
where \mu is the mean weight of grain for all the 200 piles.
Hence, the total grain weight of the population is
![\widehat{Y}=Y_1+Y_2+...+Y_{200}](https://tex.z-dn.net/?f=%5Cwidehat%7BY%7D%3DY_1%2BY_2%2B...%2BY_%7B200%7D)
=![200\times \widehat{\mu}\: \: \: =200\times 4.5\: \: \: =900\, lbs](https://tex.z-dn.net/?f=200%5Ctimes%20%5Cwidehat%7B%5Cmu%7D%5C%3A%20%5C%3A%20%5C%3A%20%3D200%5Ctimes%204.5%5C%3A%20%5C%3A%20%5C%3A%20%3D900%5C%2C%20lbs)
2) To calculate a bound on the error of estimates, we need to find the sample standard deviation.
The sample standard deviation is
S=![\sqrt{\frac{1}{5-1}\sum_{i=1}^{5}(y_i-\overline{y})^2}\: \: \: =0.9486](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cfrac%7B1%7D%7B5-1%7D%5Csum_%7Bi%3D1%7D%5E%7B5%7D%28y_i-%5Coverline%7By%7D%29%5E2%7D%5C%3A%20%5C%3A%20%5C%3A%20%3D0.9486)
Then, the standard error of ![\widehat{Y} is](https://tex.z-dn.net/?f=%5Cwidehat%7BY%7D%20is)
=83.785
Hence, a 95% bound on the error of estimates is
![[\pm z_{0.025}\times \sigma_{\widehat{Y}}]\: \: \: =[\pm 1.96\times 83.875]\: \: \: =[\pm 164.395]](https://tex.z-dn.net/?f=%5B%5Cpm%20z_%7B0.025%7D%5Ctimes%20%5Csigma_%7B%5Cwidehat%7BY%7D%7D%5D%5C%3A%20%5C%3A%20%5C%3A%20%3D%5B%5Cpm%201.96%5Ctimes%2083.875%5D%5C%3A%20%5C%3A%20%5C%3A%20%3D%5B%5Cpm%20164.395%5D)
3) Let x_1,x_2,...,x_5 denotes the total weight of the sampled piles.
Mean total weight of the sampled piles is
![\overline{x}=\frac{1}{5}\sum_{i=1}^{5}x_i=45](https://tex.z-dn.net/?f=%5Coverline%7Bx%7D%3D%5Cfrac%7B1%7D%7B5%7D%5Csum_%7Bi%3D1%7D%5E%7B5%7Dx_i%3D45)
The sample ratio is
=0.1 , this is also the estimate of the population ratio R=![\frac{\overline{Y}}{\overline{X}} .](https://tex.z-dn.net/?f=%5Cfrac%7B%5Coverline%7BY%7D%7D%7B%5Coverline%7BX%7D%7D%20.)
Therefore, the estimated total weight of grain in the population using ratio estimator is
lbs
4) The variance of the ratio estimator is
var(r)=![\frac{N-n}{N}\frac{1}{n}\frac{1}{\mu_x^2}\frac{\sum_{i=1}^{5}(y_i-rx_i)^2}{n-1} , where \mu_x=8800/200=44lbs](https://tex.z-dn.net/?f=%5Cfrac%7BN-n%7D%7BN%7D%5Cfrac%7B1%7D%7Bn%7D%5Cfrac%7B1%7D%7B%5Cmu_x%5E2%7D%5Cfrac%7B%5Csum_%7Bi%3D1%7D%5E%7B5%7D%28y_i-rx_i%29%5E2%7D%7Bn-1%7D%20%20%20%2C%20where%20%5Cmu_x%3D8800%2F200%3D44lbs)
=![\frac{200-5}{200}\, \frac{1}{5}\: \frac{1}{44^2}\, \frac{0.2}{5-1}=0.000005](https://tex.z-dn.net/?f=%5Cfrac%7B200-5%7D%7B200%7D%5C%2C%20%5Cfrac%7B1%7D%7B5%7D%5C%3A%20%5Cfrac%7B1%7D%7B44%5E2%7D%5C%2C%20%5Cfrac%7B0.2%7D%7B5-1%7D%3D0.000005)
Hence, the standard error of the estimate of the total population is
=21.556
Hence, a 95% bound on the error of estimates is
![[\pm z_{0.025}\times \sigma_{R}]\: \: \: =[\pm 1.96\times 21.556]\: \: \: =[\pm 42.25]](https://tex.z-dn.net/?f=%5B%5Cpm%20z_%7B0.025%7D%5Ctimes%20%5Csigma_%7BR%7D%5D%5C%3A%20%5C%3A%20%5C%3A%20%3D%5B%5Cpm%201.96%5Ctimes%2021.556%5D%5C%3A%20%5C%3A%20%5C%3A%20%3D%5B%5Cpm%2042.25%5D)