Question

In: Statistics and Probability

The following data give the weight (in kg) of all six employees of a company. 61...

The following data give the weight (in kg) of all six employees of a company.

61

73

53

48

68

89

  1. Construct the population distribution of the weight of the employees.

[3 marks]

  1. List all the possible samples of size four (without replacement) that can be selected from this population.

[10 marks]

  1. Calculate the mean for each of the samples in part (b).

[4 marks]

  1. Write the sampling distribution of ?̅.

[4 marks]

  1. Compute the sampling errors for the samples in part (b).

Solutions

Expert Solution

We'll answer this question using R Studio.

Since the data is weight (in kg) of employee, the variable is continuous. We will consider the distribution to be Normal with mean mu and variance sigma2.

Let X : weight of employee in kgs

a]

Using Maximum Likelihood Estimator,

mu = μ =

sigma2 = σ2 =

Code :

>w = c(61,73,53,48,68,89)
>mu = mean(w)
>mu
[1] 65.33333
>sigma2 = var(w)
>sigma2
[1] 219.4667

b]

Code :

library("utils")
samples = combn(w,4)
samples
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14] [,15]
[1,] 61 61 61 61 61 61 61 61 61 61 73 73 73 73 53
[2,] 73 73 73 73 73 73 53 53 53 48 53 53 53 48 48
[3,] 53 53 53 48 48 68 48 48 68 68 48 48 68 68 68
[4,] 48 68 89 68 89 89 68 89 89 89 68 89 89 89 89
# each column of 'samples' is a required sample

c]

Code :

# let us create a vector which will hold means of all samples
>means = c()
>for(i in 1:choose(6,4)){ means[i] = mean(samples[,i])}
>means
[1] 58.75 63.75 69.00 62.50 67.75 72.75 57.50 62.75 67.75 66.50 60.50 65.75 70.75 69.50 64.50
# means is a vector of Xbar for each sample

d]

Code :

# We need to find sampling distribution of Xbar
>Xbar_mean = mean(means)
>Xbar_mean
[1] 65.33333
>Xbar_variance = var(means)
>Xbar_variance
[1] 19.59524

Xbar follows Normal distribution with mean 65.33333 and variance 19.59524

e]

Sampling error is given by the following formula

Sampling Error = Z * sigma / (√n)

where,

Z = Z score according to confidence level , here let's take 95% of confidence level, which means (0.025, 0.975)th quantiles of Normal( 0 , 1 )

sigma = population standard deviation

n = sample size chosen.

Code :

>z_val = round(qnorm(0.975,0,1),2) # rounded off the values upto 2 decimal places
>z_val
[1] 1.96
>sigma = sqrt(sigma)
>sigma
[1] 14.81441
>n = 4

>samp_err = z_val * sigma / sqrt(n)
>samp_err
[1] 14.51812

Thus, sampling error = 14.51812


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