Question

The speeds of 80 vehicles were collected from a roadway section to estimate the mean speed. The margin of error for this sample mean was reported as 4 km/h at the confidence level of 90%. What is the sample size required to reduce this margin of error to 2 km/h (i.e., to ensure that the true mean is within ±2km/h of the sample mean 90% of the time)?

Answer 1

The margin of error is given by the equation,

Where,

E = margin of error

z = z- score corresponding to the confidence interval

σ = population standard deviation

n = sample size

Here, n = 80 (sample size greater than 30)

Thus population standard deviation (σ) = sample standard deviation (s)

Given,

E = 4

z = 1.645 (z- score corresponding to the 90 % confidence interval)

n = 80

Substituting the above values in the equation,

σ = 21.749

Here, n = 80 (sample size greater than 30)

Thus population standard deviation (σ) = sample standard deviation (s) = 21.749

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Determination of sample size for margin of error (2 kmph)

E = 2

z = 1.645 (z- score corresponding to the 90 % confidence interval)

σ = 21.749 (calculated above)

Substituting the above values in the equation,

n = 320

Thus the sample size required to reduce 4 km/h margin of error to 2 km/h = 320

Answer: Sample size required to reduce margin of error to 2 km/h is 320