Question

market, a food manufacturer uses health and diet consciousness AR the Regmentation variable. Four segments are devel- oped: 1. Concerned about cating heulthy 2. Concerned primarily about weight 3. Concerned abont health because of illness 4. Unconcerned To distinguish between groups, surveys are conducted. Ou the basis of a ques- tionnaire, people are categorized as belonging to one of these groups. A recent survey asked a random sample of 1, 250 American adults (20 and over) plete the questionnaire. The categories were recorcied. The number of adults in the sample belonging to the first gromp is 269. Find the 95% confidence intervul for the proportion of adults who ure con- cerned about cuting hcalthy foods

Answer 1

Solution:

Given the sample size, n = 1250, and number of people falling in required group, x = 269, the sample mean becomes, p^ = x/n = 269/1250 = 0.2152

Standard error = [(p^*(1 - p^))/n]1/2

SE = [(0.2152*(1 - 0.2152))/1250]1/2

SE = [(0.2152*0.7848)/1250]1/2

SE = [0.000135]1/2 = 0.0116

For 95% confidence interval, zalpha/2= z(1 - 95%)/2 = z0.025= 1.96 (using the standard normal distribution table)

Confidence interval = sample mean +/- z0.025*standard error

CI = 0.2152 +/- 1.96*0.0116

CI = 0.2152 +/- 0.0228

CI = (0.2152 - 0.0228, 0.2152 + 0.0228)

CI = (0.1924, 0.2380)