Question

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.

Answer 1

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