let the average nicotine content be 4.2 milligrams and a standard deviation - sd=1.4 milligrams.
H0 : μ=3.5 against H1 : μ>3.5
T=(X-μ)/(sd/"\\sqrt n") =1.414
table value for 1% significance and df=6 equals to 3.143>1.414, so we accept H0
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