You work for the consumer insights department of a major big

You work for the consumer insights department of a major big box retailer and you are investigating the efficacy of a new e-mail marketing campaign. Through the use of e-mail analytics research, you have determined that in a sample of 990 monitored subscribers, 258 of them opened the e-mail within 24 hours of receiving it. What is the 90% confidence interval for the true proportion of all e-mail subscribers that opened the e-mail within 24 hours of receiving it?

a) ( 0.71645 , 0.76234 )

b) ( 0.23766 , 0.28355 )

c) ( 0.24273 , 0.27849 )

d) ( 0.24665 , 0.27456 )

e) ( -0.23766 , 0.28355 )

[Please provide reasoning]

Solution

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
Mean(x)=258
Sample Size(n)=990
Sample proportion = x/n =0.26061
Confidence Interval = [ 0.26061 ±Z a/2 ( Sqrt ( 0.26061*0.73939) /990)]
= [ 0.26061 - 1.645* Sqrt(0.00019) , 0.2606 + 1.65* Sqrt(0.00019) ]
= [ 0.23766,0.28355 ]