Question

Sarah randomly selected 80 shoppers at a grocery store to taste two different types of Greek yogurt. Yogurt brand B was chosen as the favorite by 53 of them. To the nearest percent, what is the 95% confidence interval (z*-score 1.96) for the proportion of shoppers who tasted the two types of yogurt and preferred brand B? E = z* and C = + E

Answer 1

Answer:

55.89% , 76.61%

56% to 77%

Step-by-step explanation:

53 liked b

probability of B = 53/80

= 0.6625

probability of C = 1-0.6625 = 0.3375

standard error = sqrt(0.6625 * 0.3375)/80)

=0.053

So a 95% confidence interval for the population proportion is given by

0.6625 +/- 1.96*0.053

= 0.6625 +/- 0.1036

=0.5589. , 0.7661

= 55.89% , 76.61%

= 56% to 77%