In 20 packages of “Doggy Meat Snacks” there were 32 lamb snacks. The total numbe

Question

In 20 packages of “Doggy Meat Snacks” there were 32 lamb snacks. The total number of snacks in the 20 bags was 80 . We wish to calculate a 90% confidence interval for the population proportion of lamb snacks.

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=

Note: The computed L may be <0 , and the computed U may be >1 . Enter these in any case.

Explanation / Answer

Which distribution should you use for this problem?
Confidence Interval For Z - Proportion. And the data is normal

The sample proportion is p?
No. of lamb snacks(x)=32
Sample Size(n)=80
Sample proportion = x/n =0.4

The critical value zcrit of the standard normal distribution for the 90% confidence interval is?
zcrit=1.65

The standard error for p is standard error?
Standard Error = Sqrt(p*(1-p)/n))
x = Mean
n = Sample Size,
Mean(x)=32
Sample Size(n)=80
Sample proportion =0.4
Standard Error = Sqrt ( (0.4*0.6) /80) )
= 0.0548

The error bound for the proportion (EBP) is EBP ?
Margin of Error = 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
Margin of Error = Z a/2 * ( Sqrt ( (0.4*0.6) /80) )
= 1.645* Sqrt(0.003)
=0.0901

The confidence interval is given by (L,U) , where the lower bound is L?
and the upper bound is U?

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

Confidence Interval = [ 0.4 ±Z a/2 ( Sqrt ( 0.4*0.6) /80)]
= [ 0.4 - 1.645* Sqrt(0.003) , 0.4 + 1.65* Sqrt(0.003) ]
= [ 0.3099,0.4901]

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