In 15 bag of Skittle” there were 84 Red Skittle. The total number of Skittles in
ID: 3391173 • Letter: I
Question
In 15 bag of Skittle” there were 84 Red Skittle. The total number of Skittles in the 15 bags was 120. We wish to calculate a 90% confidence interval for the population proportion of red skittle.
Which distribution should you use for this problem?
The sample proportion is
p=
The critical value zcrit of the standard normal distribution for the 90% confidence interval is
zcrit=
The standard error for p is
standard error =
The error bound for the proportion (EBP) is
EBP =
The confidence interval is given by (L,U) , where
the lower bound is L= and the upper bound is U=
The computed L may be <0 , and the computed U may be >1 . Enter these in any case.
Explanation / Answer
Confidence Interval For Proportion
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)=84
Sample Size(n)=120
Sample proportion = x/n =0.7
Confidence Interval = [ 0.7 ±Z a/2 ( Sqrt ( 0.7*0.3) /120)]
= [ 0.7 - 1.645* Sqrt(0.002) , 0.7 + 1.65* Sqrt(0.002) ]
= [ 0.631,0.769]
ANS:
The sample proportion is, p=0.7
zcrit= 1.65
standard error = Sqrt ( 0.7*0.3) /120) = 0.042
The error bound for the proportion (EBP) is = 0.069
Lower = 63.1%, Upper = 76.9%