You are a travel agent and wish to estimate, with 98% confidence, the proportion of vacationers who use an online service or the Internet to make travel reservations. Your estimate must be accurate within 4% of the population proportion.
a) No preliminary estimate is available. Find the minimum sample size needed.
b) Find the minimum sample size needed, using a prior study that found 30% of the respondents said they used an online service or the Internet to make travel reservations.
c) Compare the results form parts (a) and (b).
Show steps please, I really need help I don't understand how to set this up or solve it at all!
statistics - MathGuru, Wednesday, April 6, 2011 at 6:05pm
Try this formula:
n = [(z-value)^2 * p * q]/E^2
= [(2.33)^2 * .5 * .5]/.04^2
= ? (round to the next highest whole number)
I'll let you finish the calculation.
Note: n = sample size needed; .5 for p and .5 for q are used if no value is stated in the problem. E = maximum error, which is .04 (4%) in the problem. Z-value is found using a z-table (for 98%, the value is 2.33).
Redo the problem using .3 for p and .7 for q. (Note: q = 1 - p)