In 15 bag of Skittle there were 84 Red Skittle The total num

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.

Solution

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%