Question

From generation to generation, the mean age when smokers first start to smoke varies. However, the standard deviation of that age remains constant at around 2.1 years. A survey of 37 smokers of this generation was done to see if the mean starting age is at least 19. The sample mean was 18.1 with a sample standard deviation of 1.3. Do the data support the claim at the 5% level

Answer 1

The data do not support the claim at the 5% level.

Explanation:

H0: µ ≥ 19

Ha: µ < 19

z = (X-μσ)/□(σ/(√n))

z = (18.1-19)/□(2.1/(√40)) = -2.71

Since this is a one side test the p value would be: Pv = P( z < -2.71) = 0.0033

If we compare the p value and the significance level given α = 0.05 we see that Pv < α so we can conclude that we have enough evidence to reject the null hypothesis, so in this case we have enough evidence to conclude that the true mean is significantly less than 19. At 5% level of significance, critical value is higher, and we will accept the null hypothesis. Therefore, the data does NOT support the claim at 5% level.

Answer 2

If we compare the p value and the significance level given for example
`\alpha=0.05` we see that
`p_v<\alpha` so we can conclude that we to reject the null hypothesis, and the the actual mean starting age of smokers is significantly lower than 19.

Explanation:

1) Data given and notation

`\bar X=18.1` represent the mean age when smokers first start to smoke varies

`s=1.3` represent the standard deviation for the sample

`\sigma=2.1` represent the population standard deviation

`n=37` sample size

`\mu_o =5.7` represent the value that we want to test

`\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.

We need to conduct a hypothesis in order to determine if the mean starting age is at least 19, the system of hypothesis would be:

Null hypothesis:
`\mu\geq 19`

Alternative hypothesis:
`\mu < 19`

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=(\bar X-\mu_o)/((\sigma)/(√(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".

Calculate the statistic

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

`z=(18.1-19)/((2.1)/(√(37)))=-2.607`

Calculate the P-value

Since is a one-side lower test the p value would be:

`p_v =P(z<-2.607)=0.000155`

Conclusion

If we compare the p value and the significance level given for example
`\alpha=0.05` we see that
`p_v<\alpha` so we can conclude that we to reject the null hypothesis, and the the actual mean starting age of smokers is significantly lower than 19.