Question

A cigarette industry spokesperson remarks that current levels of tar are no more than 5 milligrams per cigarette. A reporter does a quick check on 15 cigarettes representing a cross section of the market. What conclusion is reached if the sample mean is 5.63 milligrams of tar with a standard deviation of 1.61

Answer 1

Answer:

$t=\frac{5.63-5}{\frac{1.61}{\sqrt{15}}}=1.516$  

$df=n-1=15-1=14$  

Since is a right tailed test the p value would be:  

$p_v =P(t_{14}>1.516)=0.076$  

If we use a significance level of 0.05 we see that $p_v > \alpha$ and then we can conclude that we don't have evidence in order to conclude that the mean is higher than 5.63, so then the claim makes sense.

Step-by-step explanation:

Data given and notation  

$\bar X=5.63$ represent the sample mean  

$s=1.61$ represent the standard deviation for the sample  

$n=15$ sample size  

$\mu_o =15/tex] represent the value that we want to test [tex]\alpha$ represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

$p_v$ represent the p value for the test (variable of interest)  

State the null and alternative hypotheses to be tested  

We need to conduct a hypothesis in order to determine if the average is more than 5.63, the system of hypothesis would be:  

Null hypothesis:$\mu \leq 5.63$  

Alternative hypothesis:$\mu > 5.63$  

Compute the test statistic  

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

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

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

We can replace in formula (1) the info given like this:  

$t=\frac{5.63-5}{\frac{1.61}{\sqrt{15}}}=1.516$  

Now we need to find the degrees of freedom for the t distribution given by:

$df=n-1=15-1=14$  

P value

Since is a right tailed test the p value would be:  

$p_v =P(t_{14}>1.516)=0.076$  

If we use a significance level of 0.05 we see that $p_v > \alpha$ and then we can conclude that we don't have evidence in order to conclude that the mean is higher than 5.63, so then the claim makes sense.