Chapter 7. Hypothesis Testing: One-sample t-test
One-sample t-test: Test Statistic and p-value
One-sample t-test: Test Statistic
The test statistic of a one-sample #t#-test for a population mean #\mu# is denoted #t#.
The calculation of the #t#-statistic is nearly identical to the calculation of the #Z#-statistic. The only difference being:
- The #Z#-statistic is calculated using the (true) #\blue{\text{standard error of the mean}\,\sigma_{\bar{X}}}#.
\[Z = \cfrac{\bar{X}-\mu_0}{\blue{\sigma_{\bar{X}}}} = \cfrac{\bar{X}-\mu_0}{\blue{\sigma/\sqrt{n}}}\phantom{00000000000000}\] - The #t#-statistic is calculated using the #\orange{\text{estimated standard error of the mean}\,s_{\bar{X}}}#.
\[t = \cfrac{\bar{X}-\mu_0}{\orange{s_{\bar{X}}}} = \cfrac{\bar{X}-\mu_0}{\orange{s/\sqrt{n}}}\phantom{00000000000000}\]
Under the null hypothesis of a one-sample #t#-test, the #t#-statistic follows a #t#-distribution with #df = n - 1# degrees of freedom.
\[t \sim t_{n-1}\]
Student's t-Distribution
The shape of a #\boldsymbol{t}#-distribution is very similar to that of the Standard Normal Distribution, except that it has thicker tails and a lower peak. The exact shape is dependent on the number of degrees of freedom.
As the number of degrees of freedom increases, the difference between the sample standard deviation #s# and the population standard deviation #\sigma# decreases and the #t#-distribution will tend towards the Standard Normal Distribution.
Calculating the p-value of a one-sample t-test for a Population Mean with Statistical Software
The calculation of the #p#-value of a #t#-test for #\mu# is dependent on the direction of the test and can be performed using either Excel or R.
To calculate the #p#-value of a #t#-test for #\mu# in Excel, make use of one of the following commands:
\[\begin{array}{llll}
\phantom{0}\text{Direction}&\phantom{0000}H_0&\phantom{0000}H_a&\phantom{000000000}\text{Excel Command}\\
\hline
\text{Two-tailed}&H_0:\mu = \mu_0&H_a:\mu \neq \mu_0&=2 \text{ * }(1 \text{ - } \text{T.DIST}(\text{ABS}(t),n\text{ - }1,1))\\
\text{Left-tailed}&H_0:\mu \geq \mu_0&H_a:\mu \lt \mu_0&=\text{T.DIST}(t,n\text{ - }1,1)\\
\text{Right-tailed}&H_0:\mu \leq \mu_0&H_a:\mu \gt \mu_0&=1\text{ - }\text{T.DIST}(t,n\text{ - }1,1)\\
\end{array}\]
To calculate the #p#-value of a #t#-test for #\mu# in R, make use of one of the following commands:
\[\begin{array}{llll}
\phantom{0}\text{Direction}&\phantom{0000}H_0&\phantom{0000}H_a&\phantom{00000000000}\text{R Command}\\
\hline
\text{Two-tailed}&H_0:\mu = \mu_0&H_a:\mu \neq \mu_0&2 \text{ * }\text{pt}(\text{abs}(t),n\text{ - }1,lower.tail=\text{FALSE})\\
\text{Left-tailed}&H_0:\mu \geq \mu_0&H_a:\mu \lt \mu_0&\text{pt}(t,n\text{ - }1, lower.tail=\text{TRUE})\\
\text{Right-tailed}&H_0:\mu \leq \mu_0&H_a:\mu \gt \mu_0&\text{pt}(t,n\text{ - }1, lower.tail=\text{FALSE})\\
\end{array}\]
If #p \leq \alpha#, reject #H_0# and conclude #H_a#. Otherwise, do not reject #H_0#.
The engineer collects a random sample of #73# new batteries and measures the performance of each one.
She plans on using a one-sample #t#-test to determine if the mean range of the new battery significantly differs from #586# km, at the #\alpha = 0.01# level of significance.
The sample mean #\bar{X}# turns out to be #585.2# km with a standard deviation of #s=16.2# km.
Calculate the #p#-value of the test and make a decision regarding #H_0#. Round your answer to #4# decimal places.
On the basis of this #p#-value, #H_0# should not be rejected, because #\,p# #\gt# #\alpha#.
There are a number of different ways we can calculate the #p#-value of the test. Click on one of the panels to toggle a specific solution.
A sample size of #n=73# is considered large enough for the Central Limit Theorem to apply.
This means that, although the sample in question comes from a population having an unknown distribution, the test statistic
\[t=\cfrac{\bar{X} - \mu_0}{s/\sqrt{n}}\]
approximately has the #t_{n-1} = t_{72}# distribution, under the assumption that #H_0# is true.
Calculate the value of test statistic #t#:
\[t = \cfrac{\bar{X}-\mu_0}{s/\sqrt{n}} =\cfrac{585.2 - 586}{16.2/\sqrt{73}} = -0.42193\]
To calculate the #p#-value of a #t#-test, make use of the following Excel function:
T.DIST(x, deg_freedom, cumulative)
- x: The value at which you wish to evaluate the distribution function.
- deg_freedom: An integer indicating the number of degrees of freedom.
- cumulative: A logical value that determines the form of the function.
- TRUE - uses the cumulative distribution function, #\mathbb{P}(X \leq x)#
- FALSE - uses the probability density function
Since we are dealing with a two-tailed #t#-test, run the following command to calculate the #p#-value:
\[
=2 \text{ * }(1 \text{ - } \text{T.DIST}(\text{ABS}(t),n \text{ - } 1,1))\\
\downarrow\\
=2 \text{ * }(1 \text{ - } \text{T.DIST}(\text{ABS}(\text{-}0.42193), 73 \text{ - } 1,1))
\]
This gives:
\[p = 0.6743\]
Since #\,p# #\gt# #\alpha#, #H_0: \mu = 586# should not be rejected.
A sample size of #n=73# is considered large enough for the Central Limit Theorem to apply.
This means that, although the sample in question comes from a population having an unknown distribution, the test statistic
\[t=\cfrac{\bar{X} - \mu_0}{s/\sqrt{n}}\]
approximately has the #t_{n-1} = t_{72}# distribution, under the assumption that #H_0# is true.
Calculate the value of test statistic #t#:
\[t = \cfrac{\bar{X}-\mu_0}{s/\sqrt{n}} =\cfrac{585.2 - 586}{16.2/\sqrt{73}} = -0.42193\]
To calculate the #p#-value of a #t#-test, make use of the following R function:
pt(q, df, lower.tail)
- q: The value at which you wish to evaluate the distribution function.
- df: An integer indicating the number of degrees of freedom.
- lower.tail: If TRUE (default), probabilities are #\mathbb{P}(X \leq x)#, otherwise, #\mathbb{P}(X \gt x)#.
Since we are dealing with a two-tailed #t#-test, run the following command to calculate the #p#-value:
\[
2 \text{ * } \text{pt}(q = \text{abs}(t), df = n \text{ - } 1, lower.tail = \text{FALSE})\\
\downarrow\\
2\text{ * } \text{pt}(q = \text{abs}(\text{-}0.42193), df = 73 \text{ - } 1,lower.tail = \text{FALSE})
\]
This gives:
\[p = 0.6743\]
Since #\,p# #\gt# #\alpha#, #H_0: \mu = 586# should not be rejected.
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