A television sports commentator wants to estimate the proportion of citizens who "follow professional football." (a) What sample size should be obtained if he wants to be within 2 percentage points with 94% confidence if he uses an estimate of 48% obtained from a poll? The sample size is________. (Round up to the nearest integer.) (b) What sample size should be obtained if he wants to be within 2 percentage points with 94% confidence if he does not use any prior estimates? The sample size is _______. (Round up to the nearest integer.)
Accepted Solution
A:
Answer: a)2206 b) 2209Step-by-step explanation:Formula to find the sample size :a) If prior population proportion (p) is known.[tex]n=p(1-p)(\dfrac{z_{c}}{E})^2[/tex]b) If prior population proportion (p) is unknown.[tex]n=0.25(\dfrac{z_{c}}{E})^2[/tex] where, [tex]{z_{c}[/tex] is the z-value associated with confidence level and E would be the margin of error .Solution : Let p be the proportion of citizens who "follow professional football." a) Given : p=0.48Margin of error : E= 0.02z-value for 94% confidence = [tex]z_c=1.88[/tex]Required sample size would be :[tex]n=p(1-p)(\dfrac{z_{c}}{E})^2[/tex][tex]\Rightarrow\ n=0.48(1-0.48)(\dfrac{1.88}{0.02})^2[/tex]Simply , [tex]n=2205.4656\approx2206[/tex] [Rounded to the next whole number.]∴ The sample size is 2206.b) Proportion of citizens who "follow professional football" is unknown.Margin of error : E= 0.02z-value for 94% confidence = [tex]z_c=1.88[/tex]Required sample size would be :[tex]n=0.25(\dfrac{z_{c}}{E})^2[/tex][tex]\Rightarrow\ n=0.25(\dfrac{1.88}{0.02})^2[/tex]Simply , [tex]n=2209[/tex] ∴ The sample size is 2209.