Question

the claim is that smokers have a mean cotinine level greater than the level of 2.84 ng/ml found for nonsmokers (cotinine is used as a biomarker for exposure to nicotine). the sample size is ( n = 825 ) and the test statistic is ( t = 57.098 ). use technology to find the p - value. based on the result, what is the final conclusion? use a significance level of 0.01
state the null and alternative hypotheses.
( h_0:mu=2.84 )
( h_1:mu>2.84 )
(type integers or decimals. do not round.)
the test statistic is 57.10
(round to two decimal places as needed)
the p - value is (square)
(round to three decimal places as needed)

Explanation:

Step1: Determine the type of test

This is a right - tailed \(t\) - test. The formula for the \(P\) - value in a right - tailed \(t\) - test with test statistic \(t\) and degrees of freedom \(df=n - 1\) (here \(n = 825\), so \(df=825-1 = 824\)) is \(P(T>t)\) where \(T\) follows a \(t\) - distribution.

Step2: Use technology

Using statistical software (such as R: 1 - pt(57.098, 824) or Excel: =TDIST(57.098,824,1)). When the test statistic \(t = 57.098\) is extremely large, for a \(t\) - distribution (which approaches the standard normal distribution as \(df\) gets large), the \(P\) - value is very close to \(0\). Rounding to three decimal places, the \(P\) - value is \(0.000\).

Since the \(P\) - value (\(0.000\)) \(<\alpha=0.01\), we reject the null hypothesis \(H_0:\mu = 2.84\). So, there is sufficient evidence at the \(0.01\) significance level to support the claim that smokers have a mean cotinine level greater than \(2.84\) ng/mL.

Answer:

\(0.000\)