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September 21, 2014

September 21, 2014

Posted by **alex** on Wednesday, April 6, 2011 at 4:12pm.

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 6:05pmTry 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)

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