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galben [10]
3 years ago
6

Explain why increasing the sample size tends to result in a smaller sampling error when a sample mean is used to estimate a popu

lation mean.
A. The above statement is incorrect, the sample size has no effect on the sampling error.
B. The larger the sample size, the more closely the possible values of \bar{x} cluster around the mean of \bar{x}
C. If the sample size is larger, the possible values of \bar{x} are farther from the mean of \bar{x},
Mathematics
1 answer:
Anika [276]3 years ago
3 0
<span>B is the correct answer. The more values or samples you are able to add, the closer you come to approximating real life (although that is impossible) because you are more likely to have a cluster of values around the mean.</span>
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vodomira [7]

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Miguel runs 1/6 miles per minute

Step-by-step explanation:

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2 years ago
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const2013 [10]

Answer:

x = 12.48

y = 13.22

Step-by-step explanation

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10^2 + b^2 = 16^2

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3 years ago
Assume that the population proportion is 0.56. Compute the standard error of the proportion, σp, for sample sizes of 100, 200, 5
Aliun [14]

Answer:

Standard errors are 0.049, 0.035, 0.022, and 0.016.

Step-by-step explanation:

The given value of population proportion (P) = 0.56

Given sample sizes (n ) 100, 200, 500, and 1000.

Now standard error is required to calculate.

Use the below formula to find standard error.

When sample size is n = 100

\sqrt{\frac{P(1-P)}{n}} = \sqrt{\frac{0.56(1-0.56)}{100}} =0.049

When sample size is n = 200

\sqrt{\frac{P(1-P)}{n}} = \sqrt{\frac{0.56(1-0.56)}{200}} = 0.035

When sample size is n = 500

\sqrt{\frac{P(1-P)}{n}} = \sqrt{\frac{0.56(1-0.56)}{500}} =0.022

When sample size is n = 1000

\sqrt{\frac{P(1-P)}{n}} = \sqrt{\frac{0.56(1-0.56)}{1000}} = 0.016

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3 years ago
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3 0
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6.9 is the mean. you have to add them all up and divide by however many there are
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