Question

When people smoke, the nicotine they absorb is converted to cotinine, which can be measured. A sample of 60 smokers has a mean cotinine level of 172.5 ng/ml. Assuming the σ is known to be 99.5 ng/ml, test the claim that the mean cotinine level of all smokers is equal to 200.0 ng/ml. Use a 0.05 significance level.

1.   Copy and paste the Minitab output for exercise into the document underneath the problem.   You are not also required to do these by hand, unless you want to.  

2. Write the rejection rule word for word as written here, "Reject Ho if the p-value is less than or equal to alpha."

3.   Write the actual p-value and alpha, then either "Reject Ho" or "Do not reject Ho." As an example: 0.0021 ≤ 0.05. True. Reject Ho.

4.    Write an English sentence stating the conclusion, claim, and significance level. As an example: If the claim is "Can we conclude that male business executives are taller, on the average, than the general male population at the  α  = 0.05 level?", and if we found our conclusion to be do not reject Ho, the sentence would be, "There is not enough evidence to conclude that  male business executives are taller, on the average, than the general male population at the  α  = 0.05 level."

Answer 1

as instructed in question i am using minitab to solve the problem.

here, as the population sd () is given, i am using sample t test for mean.

steps:-

stat basic statistics 1 sample Z select summarized data from the drop down menu in sample size type 60,in sample mean type 172.5,in known standard deviation type 99.5tick perform hypothesis test,in hypothesized mean type 120 options in confidence level type 95,choose the alternative hypothesis as mean hypothesized meanokok.

**SOLUTION **

1).your minitab output be:-

One-Sample Z

Descriptive Statistics

N	Mean	SE Mean	95% CI for μ
60	172.5	12.8	(147.3, 197.7)

μ: mean of Sample
Known standard deviation = 99.5

Test

Null hypothesis	H₀: μ = 200
Alternative hypothesis	H₁: μ ≠ 200

Z-Value	P-Value
-2.14	0.032

2.the rejection rule:-

reject the null hypothesis if, the p-value is less than or equal to alpha.

i.e, p value   0.05 ()

3. p value = 0.032

level of significance () = 0.05

decision:-

p value =0.032   0.05 () ...REJECT H0

4).conclusion:-

there is not sufficient evidence to support the claim that the mean cotinine level of all smokers is equal to 200.0 ng/ml at 0.05 level of significance.

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