In a survey of 529 travelers 386 said that location was very

In a survey of 529 travelers, 386 said that location was very important and 323 said that room quality was very important in choosing a hotel.
a. Construct a 95% confidence interval estimate for the population proportion of travelers who said that location was very important for choosing a hotel.
b. Construct a 95% confidence interval estimate for the population proportion of travelers who said that room quality was very important in choosing a hotel.
c. Write a short summary of the information derived from (a) and (b).

Solution

a)
Confidence Interval For Proportion
CI = p ± Z a/2 Sqrt(p*(1-p)/n)))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
No.of said that location was very important (x)=386
Sample Size(n)=529
Sample proportion = x/n =0.73
Confidence Interval = [ 0.73 ±Z a/2 ( Sqrt ( 0.73*0.27) /529)]
= [ 0.73 - 1.96* Sqrt(0) , 0.73 + 1.96* Sqrt(0) ]
= [ 0.692,0.768]

b)
No..of said that room quality was very important (x)=323
Sample Size(n)=529
Sample proportion = x/n =0.611
Confidence Interval = [ 0.611 ±Z a/2 ( Sqrt ( 0.611*0.389) /529)]
= [ 0.611 - 1.96* Sqrt(0) , 0.611 + 1.96* Sqrt(0) ]
= [ 0.569,0.653]

c)
Lower the proportion narrowing the interval