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.

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

[tex]\bar X=42.6[/tex] represent the average nicotine content

[tex]s=3.7[/tex] represent the sample standard deviation

[tex]n=15[/tex] sample size

[tex]\mu_o =40[/tex] represent the value to check

[tex]\alpha=0.01[/tex] represent the significance level for the hypothesis test.

t would represent the statistic

[tex]p_v[/tex] 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:[tex]\mu \leq 40[/tex]

Alternative hypothesis:[tex]\mu > 40[/tex]

The statistic for this case would be:

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

And replcing we got:

[tex]t=\frac{42.6-40}{\frac{3.7}{\sqrt{15}}}=2.722[/tex]

The degrees of freedom are:

[tex]df=n-1=15-1=14[/tex]

The p value would be:

[tex]p_v =P(t_{(14)}>2.722)=0.0083[/tex]

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