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GuDViN [60]
3 years ago
15

How does increasing the total number of tosses from 100 to 200 (or more) affect the deviation?

Biology
1 answer:
AlekseyPX3 years ago
8 0
 <span>As a general rule, deviation decreases as you increase the sample size. </span><span>

Think of it this way. You know the average for a toin coss "should" be 50% heads, 50% tails right? 


If you only flip a coin twice, what are the chances this is going to happen? The possible results are: </span>

HH TT HT TH 


In other words, you only have a 50% chance that the outcome will be 50/50, the other two possibilities are zero heads<span>or zero tails which statistically is the complete opposite from 50/50. </span><span>

But, if you toss a coin 100, 200, or more times, you're going to continually get ever closer to that "perfect" 50% Heads, 50% Tails. By the time you've reached 1 million tosses, any deviation from "500,000 heads, 500,000 tails" would be so small that it's irrelevant. 


In other words, the deviation from the statistically perfect outcome continues to decrease the larger the sample size. 


This is why it's important to choose a reasonable sample size. Too small, and results may not be truly representative. Too large, and your results may be far more accurate than you require (i.e., it's a wasted effort). 


There are formulas for working out a viable sample size, given some possible outcomes. Check out Wikipedia's article on Sample Size Determination if you'd like further reading.</span>
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solute

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Most black bears (Ursus americanus) are black or brown in color. However, occasional white bears of this species appear in some
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Answer and Explanation:

<em><u>The number of observed individuals</u></em>:

  • AA 42
  • AG 24
  • GG 21

<u><em>Total number of individuals, N</em></u>= 87 = 42 + 24 + 21

<u><em>Allelic frequencies</em></u>:

  • f(p) = (2 x AA + AG)/ 2 x N

       f (p)= (2 x 42 + 24) /2 x 87

       f (p) = (84 + 24) / 174

       f (p)= 108 / 174

      f (p) = 0.62

  • f (q) = (2 x GG + AG)/2 x N

        f (q) = (2 x 21 + 24 )/2 x 87

        f (q) = (42 + 24)/ 174

        f (q) = 66/174

        f (q) = 0.38

p + q = 1

0.62 + 0.38 = 1

<em><u>The expected genotypic frequency:</u></em>

  • F (AA)= 0.62 ² = 0.3844
  • F (AG) = 2 x A x G = 2 x 0.62 x 0.38 = 0.4712
  • F (GG) = 0.38 ² = 0.1444

AA + AG + GG = 0.3844 + 0.4712 + 0.1444 = 1

<u><em>The number of expected individuals</em></u>:

AA= (0.62)² x 87 = 0.3844 x 87 = 33.44

AG= (0.4712) x 87 = 40.99

GG= (0.38)² x 87 = 12.563

<u><em>Total number of expected individuals</em></u> = 33.44 + 40.99 + 12.563 = 87

<u><em>Chi square</em></u>= sum (O-E)²/E

  • AA= (O-E)² /E

        AA=(42 - 33.44) ² / 33.44

        AA= 2.2

  • AB= (O-E)² /E      

        AB= (24 - 40.99)²/ 40.99

        AB=7.04

  • BB=(O-E)² /E

        BB= (21-12.563)²/12.563

        BB= 5.66

<u><em>Chi square</em></u>= sum ((O-E)²/E) = 2.2 + 7.04 + 5.66 = 14.9

<u><em>Degrees of freedom</em></u> = genotypes - alleles = 3 - 1 = 2

p value less than 0.05

There is enough evidence to reject the nule hypothesis. The genotype frequencies are not in equilibrium.

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