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 18.2 with a sample standard deviation of 1.3. Do the data support the claim at the 5% level?

Note: If you are using a Student's t-distribution for the problem, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)

1. State the distribution to use for the test. (Round your answers to four decimal places.)

X~ (__,__)

2. What is the p-value? (Round your answer to four decimal places.)

Explain what the p-value means for this problem. If:

H₀ is true, then there is a chance equal to the p-value that the average age of people when they first begin to smoke is 18.2 years or less.

H₀ is true, then there is a chance equal to the p-value that the average age of people when they first begin to smoke is not 18.2 years or less.

H₀ is false, then there is a chance equal to the p-value that the average age of people when they first begin to smoke is not 18.2 years or less.

H₀ is false, then there is a chance equal to the p-value that the average age of people when they first begin to smoke is 18.2 years or less.

3.

Construct a 95% confidence interval for the true mean. Sketch the graph of the situation. Label the point estimate and the lower and upper bounds of the confidence interval. (Round your lower and upper bounds to two decimal places.)

Answer 1

First of all, Identify H0 and HA:

H0: The mean age when smokers first start smoking is at least 19 years old. (A ≥ 19).

HA: The mean age when smokers first start smoking is younger than 19 years old. (A < 19). Set significance level: α = 0.05

1. Identify random variable and distribution for H0.

X = The average age when smokers begin smoking.

X ∼ N (19, 2.1 √ 37)

=N(19, 0.3452)

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:      

(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 = (18.2-19)/0.3452 = -2.32

seeing the z score for -2.32 from z table it is 0.4898

Calculate the P-value      

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

0.5-0.4898 = 0.0102

now since 0.0102 < 0.05 hence we do not have enough evidences to accept the null hypothesis and hence we reject the null hypothesis at 5% level.

Explainaing p values in different scinarios:

H0 is true, then there is a chance equal to the p-value that the average age of people when they first begin to smoke is 18.2 years or less.

Ans: It means our H0 defined as

H0: The mean age when smokers first start smoking is at least 19 years old. (A ≥ 19).

we have p value ≥ 0.05

H0 is true, then there is a chance equal to the p-value that the average age of people when they first begin to smoke is not 18.2 years or less.

Ans: It means our H0 defined as

H0: The mean age when smokers first start smoking is at least 19 years old. (A ≤ 19).

we have p value ≥ 0.05

H0 is false, then there is a chance equal to the p-value that the average age of people when they first begin to smoke is not 18.2 years or less.

Ans: It means our H0 defined as

H0: The mean age when smokers first start smoking is at least 19 years old. (A ≥ 19).

we have p value ≤  0.05

H0 is false, then there is a chance equal to the p-value that the average age of people when they first begin to smoke is 18.2 years or less.

Ans: It means our H0 defined as

H0: The mean age when smokers first start smoking is at least 19 years old. (A ≤ 19).

we have p value ≤  0.05

3. We have a sample mean, x = 18.2 with a sample SD of S = 2.1. For n = 37 we can assume N(18.2, 0.3452). Then using invNorm with µ = 18.2 and σ = 0.332 OR zInterval with µ = 18.2, σ = 2.1 and n = 37 We get (17.689, 18.711)