ou want to obtain a sample to estimate how much parents spend on their kids birthday parties. you believe the population standard deviation is approximately σ = 56.6 dollars.

Question

You would like to be 98% confident that your estimate is within 10 dollar(s) of average spending on the birthday parties. How many parents do you have to sample? 

n = 

Answer 1

The subject of behavior of general population is a perfect candidate of the Normal Distribution where the probabilities of infinitely large number of events converge onto a natural logarithm of the form:

N ( μ , σ) = constant * exp ( - z ² / 2)

Where,

μ: is the mean of the distribution, 10 dollars

σ: the standard deviation, 56.6 dollars

z: is the critical score (x - μ ) / σ, dimensionless parameter

constant = 1 / (σ * √2π), is a scaling parameter to normalize area under distribution curve to unity.

Therefore, the analysis starts by determining the z-score from the confidence interval of 98%.

Because, we are interested in all families that spend under or equal to 10 dollars, our z-score must account for right-tailed N-distribution. That means that an amount of half the 98%, i.e, 49%, of the right-side of N-distribution is added to the entire left half.

Thus,

P(z) = 0.5 +0.49 = 0.99

How to obtain z-score when its probability P(z) = 0.99?

We could either use simple guess based on the rule of Normal Distribution of [ 68, 95, 99 ] that corresponds to z-scores of 1, 2, and 3. Yet, due to the rough estimates of the three figures, we could come as close as saying z lies between 2 and 3.

Using a z-score calculator, we get

Z = 2.326

P(x<Z) = 0.99
P(x>Z) = 0.01
P(0<x<Z) = 0.49
P(-Z<x<Z) = 0.98
P(x<-Z or x>Z) = 0.02

Thus, with z = 2.326, we could construct the equation of the mean of the desired sample of families of sample standard deviation

σ _sample = σ / √n

OR

Estimated mean of sample = σ _sample * z

= ( σ / √n ) * z

= ( 56.6 / √n ) * 2.326

= 10 dollars

Therefore,

√n = ( 56.6 / 10 ) * 2.326

≈ 13.165

Or,

n ≈ (13.165) ²

≈ 173.3