Question

3. You frequent a popular fast food restaurant that serves tacos...and they frequently

get your order wrong. You decide to do a bit of research, so you miraculously get

ahold of n= 700 orders and realize that 242 of them were wrong. Give a 95%

confidence interval for the population proportion of incorrect orders, p.

4. If you wanted to ensure that your confidence interval had a width of at most 0.05, what is the minimum number of samples needed? Do this first assuming you have

no prior estimate for p and then under the assumption that your estimate for p from problem 3 is a reasonable one. Report both values of n.

Answer 1

Ans:

3)

sample proportion=242/700=0.3457

95% confidence interval for the population proportion

=0.3457+/-1.96*sqrt(0.3457*(1-0.3457)/700))

=0.3457+/-0.0352

=(0.3105, 0.3809)

4)

when no prior estimate,then consider p=0.5

Margin of error=CI width/2=0.05/2=0.025

Minimum number of samples needed=1.96^2*0.5*(1-0.5)/0.025^2=1537

when p=0.3457

Minimum number of samples needed=1.96^2*0.3457*(1-0.3457)/0.025^2=1390