Question

A random survey of enrollment at 10 community colleges across the United States yielded the following figures: 1,468; 1,464; 5,041; 1,450; 4,591; 5,932; 2,101; 2,029; 3,846; and 2,101. What is the 99% Error Bound (EBM) of the population mean?

Answer #1

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The answer is **3002.3 ± 1,305.241**

Solution

Count:10 (How many numbers)

Sum: 30023(All the numbers added up)

Mean:3002.3(Arithmetic mean = Sum / Count)

*Now calculate the Variance*

Sum of Differences^{2}: 25677312.1 (Add up the Squared
Differences)

Variance: 2567731.21 (Sum of Differences^{2} /
Count)

Hence Standard Deviation: 1602.414182 (The square root of the Variance)

Now,

X̄ ± Z× | σ |

√n |

Where Z is the Z-value for the chosen confidence level, X̄ is the sample mean, σ is the standard deviation, and n is the sample size. Assuming the following with a confidence level of 99%:

X̄ = 3002.3

Z = 2.576 ( as per Z table )

σ = 1602.41

n = 10

The confidence interval is:

3002.3 ± 2.576*(1602.41/√10)

i.e. 3002.3 ± 1,305.241

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