When people smoke, the nicotine they absorb is converted to cotinine, which can be measured.A random sample of 40 smokers has a mean cotinine level of 172.5 ng/ml with a sample standard deviation of 119.5 ng/ml. Use a 0.05 significance level to test the claim that the mean cotinine level of smokers is equal to 200ng/ml.
H0: μ___200 For Q23(circle one: equal to, less than , greater than, not equal to)
H1: μ___200 For Q24 (circle one:equal to, less than, greater than, not equal to)
Test Statistic t=__________ Q25 (ROUND TO 2 DECIMAL PLACES)
P-value: p =_____________ Q26 (ROUND TO4 DECIMAL PLACES)
Decision:________________________Q27 (circle one:Reject H0 , Fail to Reject H0)
Conclusion:_________________________________________________________
Q28 Circle one for the conclusion:
There is sufficient evidence to warrant the rejection of the claim.
OR
There is not sufficient evidence to warrant the rejection of the claim
Solution :
= 200
= 172.5
s =119.5
n = 405
This is the two tailed test .
The null and alternative hypothesis is ,
H0 : = 200
Ha : 200
Test statistic = t
= ( - ) / s / n
= ( 172.5- 200) /119.5 / 40
= -1.455
Test statistic = t = -1.455
P-value =0.1535
= 0.05
P-value >
0.1535 > 0.05
Fail to reject the null hypothesis .
There is not sufficient evidence to warrant the rejection of the claim
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