Question

You want to obtain a sample to estimate how much parents spend on their kids birthday parties. Based on previous study, you believe the population standard deviation is approximately

σ

=

62.5

dollars. You would like to be 95% confident that your estimate is within 3 dollar(s) of average spending on the birthday parties. How many parents do you have to sample?

Answer 1

Answer:

The number of parents spend their kids birthday parties

'n' = 1667

Step-by-step explanation:

Explanation:-

Given standard deviation of the Population = 62.5

Given the estimate error = 3 dollars

Level of significance =0.05

Z₀.₀₅ = 1.96

The margin of error defined by

$M.E = \frac{Z_{0.05} S.D}{\sqrt{n} } \\3 = \frac{1.96 X 62.5}{\sqrt{n} }$

cross multiplication , we get

3√n = 1.96 X 62.5

 √n = 40.833

Squaring on both sides, we get

n = 1667.34



The number of samples are 'n' = 1667