what is the sample mean of the 16, 21, 29, 32, 33

Question

The Neslon Company reports that children aged 2-5 years old watch about 27 hours of television weekly.  A sample of 5 children found that they watched 21, 33, 16, 29, and 32 hours of T.V. weekly.  Find the sample mean, sample variance and sample standard deviation for the number of T.V. watched.

Answer 1

Mean=sum of the hours/5=26.2 hours.

The absolute difference of each datum from the mean is the set 5.2, 6.8, 10.2, 2.8, 5.8.

And their squares are: 27.04, 46.24, 104.04, 7.84, 33.64.

The sum of these is 218.8 so variance=43.76 making the SD 6.62 approx.

[math93 is right: I should have divided 218.8 by 4 not 5: 54.7 and SD=7.40, square root of 54.7.] 

Comments

Mr.Rod He is asking for Sample variance, So if I'm right Sample variance = 54.7

Then standard deviation = 7.4

Because variance calculated by = Mean/Data count

But Sample variance calculated by = Mean/(Data count - 1)

Answer 2

a. Sample mean(x̅) (simply the arithmetic average of the sample values)

x̅ = (21+33+16+29+32)/5 = 26.2

 

b. Sample variance (s² ) (The average of the squared differences from the Mean)

s²  = [ (21-26.2)² + (33-26.2)² + (16-26.2)² + (29-26.2)² + (32-26.2)² ]/(5-1) 

      = [ (-5.2)² + (6.8)² + (-10.2)² + (2.8)² + (5.8)² ]/4

     = [ 27.04+46.24+104.04+7.84+33.64]/4

     = 218.8/4

     = 54.7

 

c. Sample standard deviation (s) (Square root of sample variance)

s = 54.7^0.5 = 7.396 ≈ 7.4

Comments

That's a good point. But I thought you only use n-1 when you're using the population mean to work out the sample mean?

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To calculate population variance use simply n. But to calculate sample variance use n-1.

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Yes, you're absolutely right! Sorry about that.