Question

# The goal is to estimate the proportion of all days in which Parsnip was fed seeds...

The goal is to estimate the proportion of all days in which Parsnip was fed seeds (as opposed to pellets) using a 99% confidence interval with margin of error no larger than 0.22. What is the minimum number of days that would need to be selected to allow the calculation of a 99% confidence interval with margin of error no larger than 0.22? It can be assumed for this problem only that the proportion of all days in which Parsnip is fed seeds is 0.58, and this number can be used for this problem.

#### Homework Answers

Answer #1

Solution :

Given that,

= 0.58

1 - = 1 - 0.58 = 0.42

margin of error = E = 0.22

At 99% confidence level the z is,

= 1 - 99%

= 1 - 0.99 = 0.01

/2 = 0.005

Z/2 = 2.58    ( Using z table )

Sample size = n = (Z/2 / E)2 * * (1 - )

= (2.58 / 0.22)2 * 0.58 * 0.42

=33.5

Sample size = 34 rounded

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