Aresearcher at a major clinic wishes to estimate the proportion of the adult population of the united states that has sleep deprivation. how large a sample is needed in orde to be 98% confident that the sample proprtion will not differ from the true proprtion by more than 5%?

The estimate of a population proportion is approximately 541.

Step-by-step explanation:

We can solve the the problem by using the formula for minimum sample needed for interval estimate of a population proportion which is given by the formula

n = pq ((Z/2) / E)^2

As, p is not defined so we use the standard p and q which is 0.5 and 0.5.

The reason for this is we have to choose form 0.1 to 0.9 both values of p and q, we will find the maximum value of pq occurs when they both are 0.5.

Next, we will find the value of (Z/2) by looking at the Z-table, we will find that at 98% confidence (Z/2) = 2.326. Now we start substituting the values in the above formula

n = (0.5)×(0.5) × (2.326/0.05)^2

n = 541.027

n ≅ 541.

439400

step-by-step explanation:

130^2 = 16900

16900*26 = 439400