A cold medicine lists milligrams of acetaminophen per fluid ounce as an active ingredient. A test of one-ounce samples of the medicine finds that the mean amount of acetaminophen for the sample is milligrams with a standard deviation of milligrams. Test the claim that the medicine does not contain the required amount of acetaminophen. Use a 0.05 significance level.

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A cold medicine lists 600 milligrams of acetaminophen per fluid ounce as an active ingredient. A test of 75 one-ounce samples of the medicine finds that the mean amount of acetaminophen for the sample is 593 milligrams with a standard deviation of 22 milligrams. Test the claim that the medicine does not contain the required amount of acetaminophen. Use a 0.05 significance level.

p-value (0.0058) < given that significance level ∝ (0.05)

So we Reject the Null Hypothesis

Step-by-step explanation:

Given that;

n = 75

x" = 593

σ = 22

H₀ : μ = 600

Hₐ : μ ≠ 600

first we find our z-score;

z = (x" - μ) / σ√n

we substitute

z = (593 - 600) / (22/√75)

= -7 / 2.5403

= - 2.76

now p-value = 2P( Z ≤ z ) { two tailed)

= 2P( Z ≤ -2.76 )

from z-table { -2.76 = 0.0029)

so

p-value = 2 × 0.0029

p-value = 0.0058

given that significance level ∝ = 0.05

p-value (0.0058) < given that significance level ∝ (0.05)

So we Reject the Null Hypothesis

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step-by-step explanation: