Question

1.

A research group conducted an extensive survey of 2896 wage and salaried workers on issues ranging from relationships with their bosses to household chores. The data were gathered through hour-long telephone interviews with a nationally representative sample. In response to the question, "What does success mean to you?" 1517 responded, "Personal satisfaction from doing a good job." Let p be the population proportion of all wage and salaried workers who would respond the same way to the stated question. Find a 90% confidence interval for p. (Round your answers to three decimal places.) 

lower limit = _______ 

upper limit = _______ 

2.

A research group conducted an extensive survey of 3019 wage and salaried workers on issues ranging from relationships with their bosses to household chores. The data were gathered through hour-long telephone interviews with a nationally representative sample. In response to the question, "What does success mean to you?" 1603 responded, "Personal satisfaction from doing a good job." Let p be the population proportion of all wage and salaried workers who would respond the same way to the stated question. How large a sample is needed it we wish to be 95% confident that the sample percentage of those equating success with personal satisfaction is within 3.0% of the population percentage? (Mint: Use P+ 0.53 as a preliminary estimate. Round your answer up to the nearest whole number.) 

_______ workers 

Answer 1

1)

sample success x =		1517
sample size n=		2896
pt estiamte p̂ =x/n=		0.5238

for 90 % CI value of z=

1.645

from excel:normsinv((1+0.9)/2)

margin of error E=z*std error   =		0.0153
lower bound=p̂ -E                       =		0.509
Upper bound=p̂ +E                     =		0.539

2)

here margin of error E =	0.030
for95% CI crtiical Z          =	1.960	from excel:normsinv((1+0.95)/2)
estimated prop.=p=	0.530
sample size n= p*(1-p)*(z/E)2=	1064	(rounding up)