Question

The mean age when smokers first start is 13 years old with a population standard deviation of 2 years. A researcher thinks that the smoking age has significantly changed since the invention of ENDS—electronic nicotine delivery systems. A survey of smokers of this generation was done to see if the mean age has changed. The sample of 30 smokers found that their mean starting age was 12.2 years old. Do the data support the claim at the 5% significance level?
What are the correct hypotheses?
H0:
H1:
Based on the hypotheses, find the following:
Test Statistic z = (Give answer to at least 4 decimal places)
Critical Values =± (Give answer to at least 4 decimal places)
Based on the above we choose to .....

a. Reject the null hypothesis
b. Accept the null hypothesis
c. Fail to reject the null hypothesis

The correct summary would be:

a. There is enough evidence to reject the claim
b. There is not enough evidence to reject the claim
c. There is enough evidence to support the claim

Answer 1

Answer:

Null hypothesis:$\mu = 13$  

Alternative hypothesis:$\mu \neq 13$

$z=\frac{12.2-13}{\frac{2}{\sqrt{30}}}=-2.191$  

a. Reject the null hypothesis

c. There is enough evidence to support the claim

Step-by-step explanation:

Data given and notation  

$\bar X=12/2$ represent the sample mean  

$s=20000$ represent the standard deviation for the sample  

$\sigma =2$ represent the population standard deviation

$n=30$ sample size  

$\mu_o =13$ represent the value that we want to test  

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

z 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 smoking age has significantly changed since the invention of ENDS—electronic nicotine delivery systems, the system of hypothesis would be:  

Null hypothesis:$\mu = 13$  

Alternative hypothesis:$\mu \neq 13$

Compute the test statistic

We know the population deviation, so for this case is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)  

z-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:  

$z=\frac{12.2-13}{\frac{2}{\sqrt{30}}}=-2.191$  

What do you conclude? Using the p-value approach  

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

$p_v =2*P(Z$  

Since the $p_v$ then the correct decision is:

a. Reject the null hypothesis

And the best conclusion is :

c. There is enough evidence to support the claim