Question

When the population distribution is normal, the statistic median {|X1 − X tilde|, . . . , |Xn − X tilde|}/0.6745 can be used to estimate σ. This estimator is more resistant to the effects of outliers (observations far from the bulk of the data) than is the sample standard deviation. Compute both the corresponding point estimate and s for the data below. (Round your answers to three decimal places.)25.01, 25.87, 26.34, 26.51, 26.75, 27.24, 27.40, 27.63, 27.83, 27.90, 28.08, 28.13, 28.37, 28.58, 28.59, 28.96, 29.20, 29.22, 29.38, 30.88

Answer 1

Answer:

The corresponding point estimate is 0.882.

The sample standard deviation is 1.373.

Step-by-step explanation:

The data set is:

S = {25.01, 25.87, 26.34, 26.51, 26.75, 27.24, 27.40, 27.63, 27.83, 27.90, 28.08, 28.13, 28.37, 28.58, 28.59, 28.96, 29.20, 29.22, 29.38, 30.88}

Compute the mean as follows:

$\bar X=\frac{1}{n}\sum X\\\\=\frac{1}{20}\times [25.01+25.87+...+30.88]\\\\=\frac{1}{20}\times 557.87\\\\=27.8935$

Subtract the mean from each value and take the modulus of those values.

The new data set is:

S₁ = {2.8835, 2.0235, 1.5535, 1.3835, 1.1435, 0.6535, 0.4935, 0.2635, 0.0635, 0.0065, 0.1865, 0.2365, 0.4765, 0.6865, 0.6965, 1.0665, 1.3065, 1.3265, 1.4865, 2.9865}

Arrange these values in ascending order as follows:

S₂ = {0.0065 , 0.0635 , 0.1865 , 0.2365 , 0.2635 , 0.4765 , 0.4935 , 0.6535 , 0.6865 , 0.6965 , 1.0665 , 1.1435 , 1.3065 , 1.3265 , 1.3835 , 1.4865 , 1.5535 , 2.0235 , 2.8835 , 2.9865}

There are 20 observations in the data set.

The median value for an even set of values is the mean of the middle two values.

In this case the median will be the mean of the 10th and 11th observations.

$\text{Median}=\frac{10^{th}obs.+11^{th}obs.}{2}=\frac{0.6965+1.0665}{2}=0.8815\approx 0.882$

Thus, the corresponding point estimate is 0.882.

Compute the standard deviation as follows:

In set S₁ we computed the absolute mean deviations.

Now take the square of these values and divide by (n - 1) to compute the sample variance:

$\sigma^{2}=\frac{1}{n-1}\sum (|X_{i}-\bar X|)^{2}$

     $=\frac{1}{20-1}\times [(2.8835)^{2}+(2.0235)^{2}+...+(2.9865)^{2}]\\\\=\frac{1}{19}\times 35.7953\\\\=1.88396$

Compute the sample standard deviation as follows:

$\sigma=\sqrt{\sigma^{2}}=\sqrt{1.88396}=1.373$

Thus, the sample standard deviation is 1.373.