Question

Consultant plans a survey to determine what % of the patients in a particular hospital were satisfied with the care they received after a major treatment. How many such patients should be surveyed so the margin of error for a 90% confidence interval is within .05? Suppose it is known from a past survey that such a % may be about 80%.

Answer 1

Solution:

Required margin of error  = 0.05

Estimated population proportion p = 0.8

Significance level = 0.10

The $\text{provided estimate population proportion}$ p is 0.8

The significance level, α = 0.1 is $z_c=1.645$, which is obtained by looking into a standard normal probability table.

The number of patients surveyed to estimate the population proportion p within the required margin of error :

$n \geq p(1-p)\left(\frac{z_c}{E}\right)^2$

  $=0.8\times (1-0.8)\left(\frac{1.64}{0.05}\right)^2$

  = 173.15

Therefore, the number of patients surveyed to satisfy the condition is n ≥ 173.15 and  it must be an integer number.

Thus we conclude that the number of patients surveyed so the margin of error of 90% confidence interval is within 0.05 are n= 174.