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ollegr [7]
3 years ago
10

A tobacco company claims that its best-selling cigarettes contain at most 40 mg of nicotine. The average nicotine content from a

simple random sample 15 cigarettes is 42.6 mg with a standard deviation (s) of 3.7 mg. Is this evidence the nicotine content of the cigarettes exceeds 40 mg? Assume cigarette nicotine content is distributed normally. Use a 1% level of significance (α=0.01) to carry out the appropriate test of significance.
Mathematics
1 answer:
tatuchka [14]3 years ago
6 0

Answer:

t=\frac{42.6-40}{\frac{3.7}{\sqrt{15}}}=2.722    

df=n-1=15-1=14  

p_v =P(t_{(14)}>2.722)=0.0083  

We see that the p value is lower than the significance level given of 0.01 so then we can conclude that the true mean for the nicotine content is significantly higher than 40 mg

Step-by-step explanation:

Information provided

\bar X=42.6 represent the average nicotine content

s=3.7 represent the sample standard deviation

n=15 sample size  

\mu_o =40 represent the value to check

\alpha=0.01 represent the significance level for the hypothesis test.  

t would represent the statistic

p_v represent the p value for the test

System of hypothesis

We want to verify if the nicotine content of the cigarettes exceeds 40 mg , the system of hypothesis are:  

Null hypothesis:\mu \leq 40  

Alternative hypothesis:\mu > 40  

The statistic for this case would be:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

And replcing we got:

t=\frac{42.6-40}{\frac{3.7}{\sqrt{15}}}=2.722    

The degrees of freedom are:

df=n-1=15-1=14  

The p value would be:

p_v =P(t_{(14)}>2.722)=0.0083  

We see that the p value is lower than the significance level given of 0.01 so then we can conclude that the true mean for the nicotine content is significantly higher than 40 mg

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