Question

Issac, BYU-Idaho student, found an old article in the school paper, the Scroll, that asserted that 11% of girls "wait for their missionary". Now that some years have passed and the missionary age limits have changed, and Isaac wants to revisit the question and would like to estimate what proportion of girls "wait for their missionary". If he is willing to accept a margin of error of 3%, and a 90% confidence level, how many observations does he need in his sample

Answer 1

Answer: 151

Step-by-step explanation:

When the prior estimate of population proportion is available , then the formula for sample size:  $n=p(1-p)(\dfrac{z^*}{E})^2$

, where p= prior estimate of population proportion

z*= critical-value.

E= Margin of  error.

Let p be the proportion of of girls "wait for their missionary".

p = 11% =0.11

E= ± 0.03

The critical z-value corresponding to 90% confidence level = z*=1.645 [using z-table ]

Substitute all the values in the above formula , we get

Required sample size  :$n=(0.11)(1-0.11)(\dfrac{(1.96)}{0.05})^2$

$\Rightarrow\ n=(0.11)(0.89)(39.2)^2$

$\Rightarrow\ n=(0.0979)(1536.64)\\\\\Rightarrow\ n=150.437056\approx151$ [Rounded to next integer.]

Thus, the number of observations he needs in his sample = 151