Question

It is reported in a news report that with 95% confidence, between 24% and 26% of New Rochelle residents went to the beach over Labor Day weekend. (a) Determine how many people were surveyed. (b) Of the people surveyed, how many responded that they went to the beach over Labor Day. (c) Do you think this would be an accurate representation of what percent of Topeka, Kansas residents went to the beach over Labor Day weekend? Explain.

Answer 1

a) The sample proportion here is computed as the middle value of the given confidence interval:

p = (0.24 + 0.26) / 2 = 0.25

The margin of error is computed as:
MOE = (0.26 - 0.24) /2 = 0.01

From standard normal tables, we have:
P( -1.96 < Z < 1.96 ) = 0.95

Therefore the sample size is computed here as:

Therefore 7204 is the required sample size here. (rounding off z value gives us 7203)

b) The number of people who responded that they went to the beach over Labor Day is computed here as:

= p*n

= 0.25*7204

= 1801

Therefore 1801 is the required number here.

c) No it wont be an accurate representation of what percent of Topeka, Kansas residents went to the beach over Labor Day weekend because we are only taking here a select sample in the sense that only one particular location cannot be generalized to other locations. There could be biases in the current sample like closeness to the beach and so on.