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A first-aid cream contains amounts of the active ingredient (in mg) that vary Normally with unknown mean m and standard deviation s from tube to tube. You select a simple random sample of 9 tubes and assess the quantity of active ingredient they contain. The 9 amounts, in mg, are 23 24 23 29 28 26 27 28 26 Based on these data, a 99% confidence interval for the population mean m has margin of error of ____ mg.

User Ddlab
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3 votes

Answer:

2.5 mg

Explanation:

Given the data:

23 24 23 29 28 26 27 28 26

The sample standard deviation, s

s = sqrt[Σ(x - mean)²] ÷ (n - 1)

Mean, m = ΣX / n

n = 9

Mean = 234 /9 = 26

Using a calculator :

s = 2.236

The margin of error, MOE

MOE = Tcritical * s/sqrt(n)

Tcritical at df = (n-1) = 9-1 = 8 , α = 0.01 = 3.35536

MOE = 3.35536 * 2.236/sqrt(9)

MOE = 3.35536 * 2.236/3

MOE = 3.35536 * 0.7453333

MOE = 2.50086

MOE = 2.5

User Ayush Singhal
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