Question

Carisoprodol, a generic muscle relaxer, claims to have, on average, at least 120 milligrams of active ingredient. An independent lab tests a random sample of 50 tablets and finds the mean content of active ingredient in this sample is 116.2 milligrams with a standard deviation of 17 milligrams. If the lab doesn't believe the manufacturer's claim, what is the approximate p-value for the suitable test

Answer 1

Answer:

The approximate p-value for the suitable test

0.05 < p < 0.1

|t| = |-1.5806| = 1.5806

t = 1.5806 < 2.009 at 0.05 level of significance

Carisoprodol, a generic muscle relaxer, claims to have, on average, is equal to 120 milligrams of active ingredient.

Step-by-step explanation:

Step(i):-

Given mean of the Population 'μ' = 120 milligrams

Given random sample size  'n' = 50

Given mean of the sample x⁻ = 116.2 milligrams

Standard deviation of the sample 'S' = 17 milligrams

Null hypothesis :  'μ' = 120

Alternative hypothesis : 'μ' < 120

Step(ii):-

Test statistic

$t = \frac{x^{-} -mean}{\frac{S}{\sqrt{n} } }$

$t = \frac{116.2 - 120}{\frac{17}{\sqrt{50} } } = \frac{-3.8}{2.404} = -1.5806$

Degrees of freedom

ν = n-1 = 50 -1 =49

t₀.₀₅ = 2.009

|t| = |-1.5806| = 1.5806

t = 1.5806 < 2.009 at 0.05 level of significance

Null hypothesis is accepted

Carisoprodol, a generic muscle relaxer, claims to have, on average, is equal to 120 milligrams of active ingredient.

P- value:-

The test statistic |t| = 1.5806 at 49 degrees of freedom

The test statistic value is lies between 0.05 to 0.1

0.05 < p < 0.1