Is the claim of 18mg of nicotine per cigarette accurate?

[tzx9633]

Homework Statement

An advertisement for a certain brand of cigarettes claimed that on average there's no more than 18mg of nicotine per cigarettes . A test of 12 cigarettes gave a sample mean of 19.1 . Assuming varience is 4 , test the claim with a significance level of α = 0.05

Homework Equations

The Attempt at a Solution

My ans is
Ho = µ_0 = 18
H1 = µ_0 <18


Since n < 30 , and standard deviation of population unknown , so , i use t-distribution

t test = (19.1-18) / ( 2 /sqrt(2) ) = 1.905

t α = 0.05 , v =11 = 1.796

Since t test > t critical , so i reject the Ho , but accroding to the ans , i should not reject the Ho ,

Is my ans wrong ?

 

[Mark44]

Homework Statement

An advertisement for a certain brand of cigarettes claimed that on average there's no more than 18mg of nicotine per cigarettes . A test of 12 cigarettes gave a sample mean of 19.1 . Assuming varience is 4 , test the claim with a significance level of α = 0.05

Homework Equations

The Attempt at a Solution

My ans is
Ho = µ_0 = 18
H1 = µ_0 <18


Since n < 30 , and standard deviation of population unknown , so , i use t-distribution

t test = (19.1-18) / ( 2 /sqrt(2) ) = 1.905

t α = 0.05 , v =11 = 1.796

Since t test > t critical , so i reject the Ho , but accroding to the ans , i should not reject the Ho ,

Is my ans wrong ?

Your null hypothesis should be $H_0: \mu_0 \le 18$
In addition, your calculation of the t statistic looks wrong to me.

In testing the null hypothesis that the population mean is equal to a specified value μ0, one uses the statistic

where

is the sample mean, s is the sample standard deviation of the sample and n is the sample size.

 

[Ray Vickson]

Homework Statement

An advertisement for a certain brand of cigarettes claimed that on average there's no more than 18mg of nicotine per cigarettes . A test of 12 cigarettes gave a sample mean of 19.1 . Assuming varience is 4 , test the claim with a significance level of α = 0.05

Homework Equations

The Attempt at a Solution

My ans is
Ho = µ_0 = 18
H1 = µ_0 <18


Since n < 30 , and standard deviation of population unknown , so , i use t-distribution

t test = (19.1-18) / ( 2 /sqrt(2) ) = 1.905

t α = 0.05 , v =11 = 1.796

Since t test > t critical , so i reject the Ho , but accroding to the ans , i should not reject the Ho ,

Is my ans wrong ?

You want to distinguish between values of $\mu$ that are $\leq 18$ and $> 18$, so you should use $H_0: \mu= 18$ vs. $H_1: \mu > 18$.

Are you sure you should use $\sigma = 2?$ If this refers to the sample variance, it is the estimated variance of $X$, not of $\bar{X}_{12}$ = sample mean. The appropriate standard deviation for $\bar{X}_{12}$ is $\sigma/\sqrt{12} = 2/\sqrt{12}.$ That would give an even larger value of $t$ than the one you used, making $H_0$ even more strongly rejected than you indicate. This seems to be another instance of a wrong answer in the book (or possibly, the book having a weird, non-standard way of doing things).