Question

Real Fruit Snacks” there were five Bam-Bam snack pieces. The total number of snack pieces in the six bags was 68. We wish to calculate a 96% confidence interval for the population proportion of Bam-Bam snack pieces. Construct a 96% confidence interval for the population proportion of Bam-Bam snack pieces per bag and calculate the error bound.

Do you think that six packages of fruit snacks yield enough data to give accurate results? Why or why not?

Answer 1

solution:

given x=5

n=68

sample proportion=p^=x/n=5/68= 0.07352941

z crit for 96%==NORM.S.INV(0.98)=2.053749

96% confidence interval for proportion is given by

p^-Z*qrt(p^*(1-p^/n),p^+Z*qrt(p^*(1-p^/n)

0.07352941-2.053749*sqrt(0.07352941*(1-0.07352941)/68), 0.07352941+2.053749*sqrt(0.07352941*(1-0.07352941)/68)

0.008525532, 0.1385333

we are 96% confident that the true population proportion of Bam Bam snake pieces per bag lies in between

0.00852553 and , 0.1385333

Error bound is  0.06500388

np^=68*5/68=5<10

n(1-p^)=68*(1-5/68)=63>10

Both np and nq>10,here np<5

conditions for using normal distribution is not satisfied

please give me thumb up