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The size of the sample should not affect the critical value.

Q: What happens to the crtitical value for a chi square test if the size of the sample is increased?

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The area is increased by a factor of 9.

The area is increased to 3x3=9 times the original.

10

This is done by measuring a sample of the item and weighing it. It is easiest if the sample measures 1 square metre, or an exact multiple of it.

2a2+2a2 = 4a2

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The area is increased by a factor of 9.

The area is increased by a factor of 9.

The size of the force decreases. It is inversely proportional to the square of the distance.

The area is increased to 3x3=9 times the original.

The area is increased to 22 = 4 times what is was.The area is increased to 22 = 4 times what is was.The area is increased to 22 = 4 times what is was.The area is increased to 22 = 4 times what is was.

10

Well, sort of. The Chi-square distribution is the sampling distribution of the variance. It is derived based on a random sample. A perfect random sample is where any value in the sample has any relationship to any other value. I would say that if the Chi-square distribution is used, then every effort should be made to make the sample as random as possible. I would also say that if the Chi-square distribution is used and the sample is clearly not a random sample, then improper conclusions may be reached.

The standard error should decrease as the sample size increases. For larger samples, the standard error is inversely proportional to the square root of the sample size.The standard error should decrease as the sample size increases. For larger samples, the standard error is inversely proportional to the square root of the sample size.The standard error should decrease as the sample size increases. For larger samples, the standard error is inversely proportional to the square root of the sample size.The standard error should decrease as the sample size increases. For larger samples, the standard error is inversely proportional to the square root of the sample size.

This is done by measuring a sample of the item and weighing it. It is easiest if the sample measures 1 square metre, or an exact multiple of it.

The sample standard deviation (s) divided by the square root of the number of observations in the sample (n).

square, rectangle

2a2+2a2 = 4a2