Answer to Question #196172 in Statistics and Probability for Ichigo

Question #196172

Measurements of a sample of 6 weights were found to be 14⋅ 16,3 ⋅ 15,6 ⋅ 14,7 ⋅ 16,8 ⋅ 2

and 15 ⋅ 4 kilogram respectively. (i) Determine an unbiased estimate of the population 

mean. (ii) Compare the sample standard deviation with the estimated population 

standard deviation. 


1
Expert's answer
2021-05-25T13:34:52-0400
14,14.7,15,15.6,16.3,16.814, 14.7, 15, 15.6, 16.3, 16.8


(i)

Arithmetic mean xˉ=i=16xi6=92.46=15.4\bar{x}=\dfrac{\displaystyle\sum_{i=1}^{6}x_i}{6}=\dfrac{92.4}{6}=15.4


Median = 15+15.62=15.3\dfrac{15+15.6}{2}=15.3


Mode: All values appeared just once.

The sample mean xˉ=15.4,\bar{x}=15.4, is an unbiased estimator of the population mean.


(i)  Variance

The variance of a sample is:


s2=161i=16(xixˉ)2=5.425=1.084s^2=\dfrac{1}{6-1}\displaystyle\sum_{i=1}^{6}(x_i-\bar{x})^2=\dfrac{5.42}{5}=1.084


 Standard Deviation


s=s2=1.0841.041153s=\sqrt{s^2}=\sqrt{1.084}\approx1.041153




σ2=16i=16(xiμ)2=5.426=2.713\sigma^2=\dfrac{1}{6}\displaystyle\sum_{i=1}^{6}(x_i-\mu)^2=\dfrac{5.42}{6}=\dfrac{2.71}{3}

σ=σ2=2.7130.950438<s\sigma=\sqrt{\sigma^2}=\sqrt{\dfrac{2.71}{3}}\approx0.950438<s


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