In a survey of 500 U.S. adults
- In a survey of 500 U.S. adults, 45% of them said that loungingat the beach was their “dream vacation.’ Assuming this to be asimple random sample of U.S. adults, the lower confidencelimit of the 95% confidence intervals for the proportionof U.S. adults who consider lounging at the beach to be their dreamvacation would equal to;
Answer:
Solution :
n = 500
= 45% = 0.45
1 –= 1 – 0.45 = 0.55
At 95% confidence level the z is ,
=1 – 95% = 1 – 0.95 = 0.05
/ 2 = 0.05 / 2 = 0.025
Z/2= Z0.025 = 1.960
Margin of error = E = Z
/ 2 * 
* (1 – )) / n)
= 1.960 * (
= 0.044
At 95% confidence interval estimate of the population meanis,
– E < 
< + E
(0.450 – 0.044 < < 0.450 + 0.044 )
0.406 < < 499
lower confidence limit = 0.406
upper confidence limit = 0.499
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