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Answer:
The approximate p-value for the suitable test
<em>0.05 < p < 0.1</em>
|t| = |-1.5806| = 1.5806
t = 1.5806 < 2.009 at 0.05 level of significance
<em>Carisoprodol, a generic muscle relaxer, claims to have, on average, is equal to 120 milligrams of active ingredient.</em>
<u>Step-by-step explanation</u>:
<u><em>Step(i)</em></u>:-
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
<u><em>Step(ii):</em></u>-
<em>Test statistic </em>
<em>Degrees of freedom</em>
<em>ν = n-1 = 50 -1 =49</em>
t₀.₀₅ = 2.009
|t| = |-1.5806| = 1.5806
t = 1.5806 < 2.009 at 0.05 level of significance
<em>Null hypothesis is accepted</em>
<em>Carisoprodol, a generic muscle relaxer, claims to have, on average, is equal to 120 milligrams of active ingredient.</em>
<u><em>P- value:-</em></u>
<em>The test statistic |t| = 1.5806 at 49 degrees of freedom</em>
<em>The test statistic value is lies between 0.05 to 0.1</em>
<em>0.05 < p < 0.1</em>
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