A survey of 600 residents reports 261 of them support a tax levy to build a new

ID: 3151310 • Letter: A

Question

firehouse. (a) Find a 90% confidence interval for the actual proportion of residents

who support the levy. (b) State the assumptions used in your analysis. (c) If this is

a preliminary study and we wish to find a 90% confidence interval for the proportion

of residents who support the levy that has a margin of error less than .03, how many

people should be sampled? (d) If no preliminary study had been done and we wished to

find a 90% confidence interval for the actual proportion of residents who support the

levy with a margin of error less than .03, how many people should be sampled?

alpha = 0.10 alpha / 2 = 0.05 Z = 1.64

p = 261 /600 = 0.435

0.435 +/- 1.64 * srqt ( 0.435*0.565 / 600 )

0.435 +/- 0.033

0.402 < p < 0.468

this is a normal probability function , because we are talking about proportions

alpha / 2 = 0.05 Z=1.64

n = 1.64^2 * 0.435*0.565 / 0.03^2

n = 734.48 = 735

n = 1.64^2 * 0.5 * 0.5 / 0.03^2

n = 747.11

n = 748

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