Question

The magic dragon cigarette company claims that their cigarettes contain an average of only 10 mg of tar. a random sample of 100 magic dragon cigarettes shows the average tar content to be 11.5 mg with a known population standard deviation of 4.5 mg. construct a hypothesis test to determine whether the average tar content of magic dragon cigarettes exceeds 10 mg. use a 5% level of significance.
what is the critical value method and pvalue method

Answer 1

Answer:

$t=\frac{11.5-10}{\frac{4.5}{\sqrt{100}}}=3.333$    

Critical value

The significance is 5% so then $\alpha=0.05$ and $\alpha/2=0.025$ then the critical value for this case would be $z_{\alpha/2}= 1.64$. Since the calculated value is higher than the critical value we have enough evidence to reject the null hypothesis and we can conclude that the true mean is higher than 10 mg

P value

The p value would be given

$p_v =P(z>3.333)=0.000434$  

Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly higher than 10 mg

Step-by-step explanation:

Information given

$\bar X=11.5$ represent the sample mean

$\sigma=4.5$ represent the population standard deviation

$n=100$ sample size  

$\mu_o =10$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.  

t would represent the statistic

$p_v$ represent the p value

Hypothesis to test

We want to test if the true mean is higher than 10mg, the system of hypothesis would be:  

Null hypothesis:$\mu \leq 10$  

Alternative hypothesis:$\mu > 10$  

the statistic for this case would be given by:

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

Replacing we got:

$t=\frac{11.5-10}{\frac{4.5}{\sqrt{100}}}=3.333$    

Critical value

The significance is 5% so then $\alpha=0.05$ and $\alpha/2=0.025$ then the critical value for this case would be $z_{\alpha/2}= 1.64$. Since the calculated value is higher than the critical value we have enough evidence to reject the null hypothesis and we can conclude that the true mean is higher than 10 mg

P value

The p value would be given

$p_v =P(z>3.333)=0.000434$  

Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly higher than 10 mg