Question

A marketer wants to determine the proportion of London shoppers who have purchased gifts online throughout the year. From a random sample of 2,079 consumers, 1,396 consumers stated that they purchased their holiday gifts online last season. (Round all final statistics and final answer to fourdecimal places)

(a) Calculate and construct, the 95% confidence interval estimate for the proportion of London consumers who purchased their gifts online. (3 points)

(b) INTERPRET YOUR RESULTS

(c) Is it reasonable to say that 70% of London Shoppers purchase gifts online? EXPLAIN! (1 point)

Answer 1

n = Total number of consumers in a random sample = 2079

x = number of consumers stated that they purchased their holiday gifts online last season = 1396

Sample proportion:

(Round to 4 decimal)

a)

Confidence level = c = 0.95

95% confidence interval estimate for the proportion of London consumers who purchased their gifts online is

where zc is z critical value for (1+c)/2 = (1+0.95)/2 = 0.975

zc = 1.96 (From statistical table of z values)

   (Round to 4 decimal)

95% confidence interval estimate for the proportion of London consumers who purchased their gifts online is (0.6513, 0.6917)

b)

Interpretation:

We are 95% confident that the true proportion of London consumers who purchased their gifts online will lie in the interval (0.6513, 0.6917).

c)

Here confidence interval does not contain 0.70

So it is not reasonable to say that 70% of London Shoppers purchase gifts online.