Respuesta :
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
[tex]t = \frac{x^{-} -mean}{\frac{S}{\sqrt{n} } }[/tex]
[tex]t = \frac{116.2 - 120}{\frac{17}{\sqrt{50} } } = \frac{-3.8}{2.404} = -1.5806[/tex]
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
