Question

For a population, the mean is 19.4 and the standard deviation is 5.8. Compare the mean and standard deviation of the following random samples to the population parameters.

25, 32, 16, 12, 11, 38, 22, 21, 19, 20

Answer 1

Answer:

The mean of sample (21.6) is greater than population mean and the standard deviation of sample (8.4) is also greater than population standard deviation.

Both values, mean and std are greater than corresponding population values. This means, that our sample overestimates or is skewed to the right in the values of population. Also each specific value is farther apart from the mean so variation is much bigger in our sample than in our population.

Step-by-step explanation:

To calculate the mean, we estimate the avergage of the data as

Mean = Sum(data)/n

Where n is the number of observations of sample. We have;

(25 + 32 +  16 + 12 + 11 + 38 + 22 + 21  + 19 + 20)/10 = 21.6

The standard deviation is the square root of variance. Variance is sum of the deviation of each data point to the sample mean.

Variance = sum i= 1 to n (Xi-xmean)/(n-1)

Where n = 10 the # of observations and xi is each specific data point.

If you calculate it in Excel or or by hand you obtain:

Variance = sum i= 1 to n (Xi-21.6)/(10-1) = 8.4

Both values, mean and std are greater than corresponding population values. This means, that our sample overestimates or is skewed to the right in the values of population. Also each specific value is farther apart from the mean so variation is much bigger in our sample than in our population.

Answer 2

Answer:

The mean of the random samples (21.6) is greater than population´s mean (19.4) and the STD of the random samples (7.96) is greater than population´s STD (5.8)

Step-by-step explanation:

To calculate the mean of the random samples, we find the average value of the data set

$Mean=\frac{(\sum_{i=0}^{n}{a_i})}{n}$

Where a is an element of the random example and n is the number of elements in the sample, as follows

Mean = (25+32+16+12+11+38+22+21+19+20)*(1/10) = 21.6

To calculate the STD (Standard Deviation) we need to know the variance, because

$STD=\sqrt{var}$

The variance is determined as the sum of the deviation from one element of the data to the mean squared, all divided by n as below.

$Var=(\sum_{i=1}^{n}{(a_{i}-Mean)^{2})/n$

If you calculate this by calculator or with a computer, you should get Var=63.44

And with that value, STD=7.96≈8

Therefore, by comparison, both Mean and STD of the random sample are greater than the population´s parameter