Question

1) A box contains 10,000 tickets: 4,000 0's and 6,000 1's. And 10,000 draws will be made at random with replacement from this box. Which of the following best describes the situation, and why?
i. The number 1's will be 6,000 exactly
ii. The number of 1's is very likely to equal 6,000, but there is also some small chance that it will not be equal to 6,000
iii. The number of 1's is likely to be different from 6,000, but the difference is likely to be small compared to 10,000

2) Repeat question one for 1 for 10,000 draws made at random without replacement from the box? Why?

3) 50 draws are made at random with replacement from the box
1 2 3 4 5; the sum of the draws turns out to be 157.
What is the expected value for the sum? Why?
What is the observed value? Why?
What is the chance error? Why?
What is the standard error? Why?

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3)

Expected number in each draw = average of 1,2,3,4,5 = 3
So,
Expected sum in 50 draws = 50 *3 =150
Observed value = 157 because of variance in random sampling method.
Chance error = 157 -150 = 7 due to variance in random sampling menthod.
Standard error for single draw = standard deviation of 1,2,3,4,5 i.e. 1.414 (using stdevp function in Excel)
So, variance = 50 *1.4142=100
Therefore, standard error = square root of variance = 10...

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