Question

In each of parts​ (a)-(c), we have given a likely range for the observed value of a sample proportion p. Based on the given​ range, identify the educated guess that should be used for the observed value of p to calculate the required sample size for a prescribed confidence level and margin of error.

a. 0.2 to 0.3

b. 0.1 or less

c. 0.3 or greater

Answer 1

SOLUTION:

From given data,

calculate the required sample size for a prescribed confidence level and margin of error.

(a) 0.2 to 0.3

Given the confidence interval for proportion as 0.2 and 0.3

Then the sample proportion = p = (Upper limit + lower limit) /2

= (0.2+0.3)/2

=  0.25

q = 1-p = 1-0.25 = 0.75

The margin of error = Z_/2 * sqrt( pq / n)

[Assume the level of significance is assumed to be 5% then ,Z_/2 = 1.96 ]

Then, for the sample proportion, margin of error is

E =1.96 * sqrt( 0.25*0.75 / n)

E = 0.8487 / sqrt(n)

b. 0.1 or less

Given the confidence interval for proportion as 0.1 or less

Guess the lower limit as 0

Then the sample proportion = p = (Upper limit + lower limit) /2

= (0.1+0)/2

=  0.05

q = 1-p = 1-0.05= 0.95

The margin of error = Z_/2 * sqrt( pq / n)

[Assume the level of significance is assumed to be 5% then ,Z_/2 = 1.96 ]

Then, for the sample proportion, margin of error is

E =1.96 * sqrt( 0.05*0.95/ n)

E = 0.4272 / sqrt(n)

c. 0.3 or greater

Given the confidence interval for proportion as 0.3 or greater

Guess the upper limit as 1

Then the sample proportion = p = (Upper limit + lower limit) /2

= (1+0.3)/2

=  0.65

q = 1-p = 1-0.65= 0.35

The margin of error = Z_/2 * sqrt( pq / n)

[Assume the level of significance is assumed to be 5% then ,Z_/2 = 1.96 ]

Then, for the sample proportion, margin of error is

E =1.96 * sqrt( 0.65*0.35/ n)

E = 0.9349 / sqrt(n)

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