A soft drink manufacturer purchases aluminum cans from an outside vendor. A rand

Question

A soft drink manufacturer purchases aluminum cans from an outside vendor. A random sample of 66 cans is selected from a large shipment, and each is tested for strength by applying an increasing load to the side of the can until it punctures. Of the 66 cans, 54 meet the specification for puncture resistance. Find a 90% confidence interval for the proportion of cans in the shipment that meet the specification.

Carry four significant digits throughout your calculations.

Lower confidence bound =

Upper confidence bound =

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)=54
Sample Size(n)=66
Sample proportion = x/n =0.8182
Confidence Interval = [ 0.8182 ±Z a/2 ( Sqrt ( 0.8182*0.1818) /66)]
= [ 0.8182 - 1.64* Sqrt(0.0023) , 0.8182 + 1.64* Sqrt(0.0023) ]
= [ 0.7403,0.8961]

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