 Mathematics
Large Sample Estimation of a Population Proportion

### Question Description  a) The best point estimate is the sample proportion, which is:

822/1250 = 0.6576

b) Since we don't know the population standard deviation, only that for the sample, we should use a t-distribution here, with df = n-1 = 1249. The critical t-stat for a 98% confidence interval will be:

t_crit = t.inv(p=0.99,df=1249) = 2.329337192

The sample standard deviation is given by sqrt(p(1-p)) = sqrt(0.6576(1-0.6576))

0.474512634

The standard error is sample sd / sqrt(n) = 0.474512634/sqrt(1250)

SEM = 0.013421244

So the 98% confidence interval for the sample mean is given by

sample mean +/- t_crit x SEM

= 0.6576 +/- 2.329337192 x 0.013421244

= [0.6263 to 0.6889] iainharlow (1041)
Rice University Anonymous
Top quality work from this tutor! I’ll be back! Anonymous
Just what I needed… fantastic! Anonymous
Use Studypool every time I am stuck with an assignment I need guidance. Studypool 4.7 Trustpilot 4.5 Sitejabber 4.4 Brown University

1271 Tutors California Institute of Technology

2131 Tutors Carnegie Mellon University

982 Tutors Columbia University

1256 Tutors Dartmouth University

2113 Tutors Emory University

2279 Tutors Harvard University

599 Tutors Massachusetts Institute of Technology

2319 Tutors New York University

1645 Tutors Notre Dam University

1911 Tutors Oklahoma University

2122 Tutors Pennsylvania State University

932 Tutors Princeton University

1211 Tutors Stanford University

983 Tutors University of California

1282 Tutors Oxford University

123 Tutors Yale University

2325 Tutors