A researcher wishes to estimate the percentage of adults who support abolishing the penny. What size sample should be obtained if he wishes the estimate to be within 3 percentage points with 99?% confidence if

?(a) he uses a previous estimate of 34?%?

?(b) he does not use any prior? estimates?

?(a) n=

?(b) n=

a)

?1−α=0.99E=0.03p^?=0.34?(Confidence level).(Error).(Initial estimate of p)??Sample size.An estimate p^? from a previous sample is available: n=(Ezα/2??)2 p^? (1−p^?)??Calculus of zα/2?− value.1−α=0.99α=1−0.99α=0.01α/2=20.01?α/2=0.0050zα/2?−value is the z−value having an area of α/2 (0.0050) to the right. The cumulative area to the left is 1−α/2=1−0.0050=0.9950?Calculus of zα/2? using the cumulative standard normal distribution table.We search through the probabilities to find the value that corresponds to 0.9950.−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−??????????z...2.32.42.52.62.7...?0.00...0.98930.99180.99380.99530.9965...?0.01...0.98960.99200.99400.99550.9966...?0.02...0.98980.99220.99410.99560.9967...?0.03...0.99010.99250.99430.99570.9968...?0.04...0.99040.99270.99450.99590.9969...?0.05...0.99060.99290.99460.99600.9970...?0.06...0.99090.99310.99480.99610.9971...?0.07...0.99110.99320.99490.99620.9972...?0.08...0.99130.99340.99510.99630.9973...?0.09...0.99160.99360.99520.99640.9974...??????????−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−?We do not find 0.9950 exactly; the nearest value is corresponding to 0.9951 therefore:zα/2?=2.5+0.08zα/2?=2.58???Since we know an estimate p^? from a previous sample (p^?=0.34):n=(0.032.58?)2∗0.34∗(1−0.34)n=(86)2∗0.34∗0.66n=7396∗0.2244n=1659.6624nmin?=1660The minimum sample size is 1660.???

b)

?1−α=0.99E=0.03?(Confidence level).(Error).??Sample size.An estimate p^? from a previous sample is not available: n=(Ezα/2??)2 (0.25)??Calculus of zα/2?− value.1−α=0.99α=1−0.99α=0.01α/2=20.01?α/2=0.0050zα/2?−value is the z−value having an area of α/2 (0.0050) to the right. The cumulative area to the left is 1−α/2=1−0.0050=0.9950?Calculus of zα/2? using the cumulative standard normal distribution table.We search through the probabilities to find the value that corresponds to 0.9950.−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−??????????z...2.32.42.52.62.7...?0.00...0.98930.99180.99380.99530.9965...?0.01...0.98960.99200.99400.99550.9966...?0.02...0.98980.99220.99410.99560.9967...?0.03...0.99010.99250.99430.99570.9968...?0.04...0.99040.99270.99450.99590.9969...?0.05...0.99060.99290.99460.99600.9970...?0.06...0.99090.99310.99480.99610.9971...?0.07...0.99110.99320.99490.99620.9972...?0.08...0.99130.99340.99510.99630.9973...?0.09...0.99160.99360.99520.99640.9974...??????????−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−?We do not find 0.9950 exactly; the nearest value is corresponding to 0.9951 therefore:zα/2?=2.5+0.08zα/2?=2.58???Since we do not know an estimate p^? from a previous sample:n=(0.032.58?)2∗0.25n=(86)2∗0.25n=7396∗0.25n=1849nmin?=1849The minimum sample size is 1849.??