Question

A wheat farmer cuts down the stalks of wheat and gathers them in 200 piles. The 200 gathered piles will be put on a truck. The truck goes to a large flour manufacturer who will separate the grain kernels from the stalks, and just pay the farmer for the grain. The farmer wants to estimate the total weight of grain on the field, and draws a simple random sample of 5 piles from the 200 piles on the field. For each pile sampled the farmer weighs the pile, then separates the grain kernels from the stalks, and weighs the grain kernels. The sample data are as follows (weight is in lbs)

Pile 1 Pile 2 Pile 3 Pile 4 Pile 5
Weight of pile 30 40 50 60 45
Weight of grain in pile 3.3 4.1 4.7 5.9 4.4
1. Estimate the total weight of grain in all 200 piles
2. Give a bound on the error of estimation in Question1.

Answer 1

1.900lbs

2. + or - 164.395

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 size

are N=200 & n=5 respectively.

1.Let Y1,Y2,...,Y{200} be the weight of grain in the 200 piles and y1,y2,...,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,

Therefore,

Mean= (3.3+4.1+4.7+5.9+4.5)/5=4.5

where \mu is the mean weight of grain for all the 200 piles.

Hence, the total grain weight of the population is

Y=Y1+Y2+...+Y{200}

=200×mean: =200× 4.5= 900lbs

2.

To calculate a bound on the error of estimates, we need to find the sample standard deviation.

The sample standard deviation is= 0.9486

And the standard error=83.785

Hence, a 95% bound on the error of estimates is

error of estimates is + or - 164.395

The formula for standard deviation and standard error is attached

Answer 2

Check the explanation

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}=(1)/(5)\sum_(i=1)^(5)y_i=(1)/(5)(3.3+4.1+4.7+5.9+4.5)=4.5`

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)`

=
`200* \widehat{\mu}\: \: \: =200* 4.5\: \: \: =900\, lbs`

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{(1)/(5-1)\sum_(i=1)^(5)(y_i-\overline{y})^2}\: \: \: =0.9486`

Then, the standard error of
`\widehat{Y} is`

`\sigma_{\widehat{Y}} =\sqrt{(N^2S^2)/(n)\bigg((N-n)/(N)\bigg)}\: \: \:`=83.785

Hence, a 95% bound on the error of estimates is

`[\pm z_(0.025)* \sigma_{\widehat{Y}}]\: \: \: =[\pm 1.96* 83.875]\: \: \: =[\pm 164.395]`

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}=(1)/(5)\sum_(i=1)^(5)x_i=45`

The sample ratio is

`r=\frac{\overline{y}}{\overline{x}}=(4.5)/(45)`=0.1 , this is also the estimate of the population ratio R=
`\frac{\overline{Y}}{\overline{X}} .`

Therefore, the estimated total weight of grain in the population using ratio estimator is

`\widehat{Y}_R\: \: =r* 8800\: \: =0.1* 8800\: \: =880\,` lbs

4) The variance of the ratio estimator is

var(r)=
`(N-n)/(N)(1)/(n)(1)/(\mu_x^2)(\sum_(i=1)^(5)(y_i-rx_i)^2)/(n-1) , where \mu_x=8800/200=44lbs`

=
`(200-5)/(200)\, (1)/(5)\: (1)/(44^2)\, (0.2)/(5-1)=0.000005`

Hence, the standard error of the estimate of the total population is

`\sigma_R=√(X^2 \: var(r))\: \: \: =√(8800^2* 0.000005)\: \: \:`=21.556

Hence, a 95% bound on the error of estimates is

`[\pm z_(0.025)* \sigma_(R)]\: \: \: =[\pm 1.96* 21.556]\: \: \: =[\pm 42.25]`