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Student[Statistics][OneSampleTTest] Overview

overview of the One Sample T-Test Description

 • One Sample T Test is used to test if the sample studied follows a normal distribution when its standard deviation is unknown and the mean is known. The mean is equal to the test value based on the sample drawn.
 If the standard deviation is known for the assumed normal distribution, One Sample Z Test is used instead.
 • Requirements for using One Sample T Test:
 1 The sample studied is assumed to follow a normal distribution.
 2 The standard deviation of the assumed normal distribution is unknown.
 • The formula is:

$T=\frac{\left(\mathrm{Mean}\left(X\right)-{\mathrm{\mu }}_{0}\right)\sqrt{N}}{s}$

 where $X$ is the sample, ${\mathrm{\mu }}_{0}$ is the test value of mean, $s$ is the sample standard deviation, $N$ is the sample size, and $T$ follows Student's T distribution with $N-1$ degrees of freedom. Example

A research team did a survey on male smokers over 20 in Ontario to figure out the age when they started to smoke. Knowing that the starting age is normally distributed, a statistical test was run to test whether the average starting age is 15. 1000 male smokers were randomly selected and interviewed. The result shows that the sample standard deviation of the 1000 smokers is 5.144 and their average age of beginning to smoke is 16.

 1 Determine the null hypothesis:
 Null Hypothesis: ${\mathrm{\mu }}_{0}=15$ (the actual mean)
 2 Substitute the information into the formula:
 $t=\frac{\left(16-15\right)}{\left(\frac{5.144}{\sqrt{1000}}\right)}=6.1475$
 3 Compute the p-value:
 $p-\mathrm{value}=\mathrm{Probability}\left(|T|>6.1475\right)=\mathrm{Probability}\left(T<-6.1475\right)+\mathrm{Probability}\left(T>6.1475\right)=p=1.136158\cdot {10}^{-9}$, $T˜\mathrm{StudentT}\left(999\right)$
 4 Draw the conclusion:
 This statistical test provides evidence that the null hypothesis is false, so we reject the null hypothesis.