One-sample t-test: Test Statistic and p-value

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.

Student's t vs. Standard Normal Distribution

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 $p \lt \alpha$ , reject $H_0$ and conclude $H_a$ . Otherwise, do not reject $H_0$ .

An engineer at Tesla is working on a new type of lithium-ion battery. It is known that cars with the old battery have an average range of $587$ km when the battery is fully charged.

The engineer collects a random sample of $56$ 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 $587$ km, at the $\alpha = 0.07$ level of significance.

The sample mean $\bar{X}$ turns out to be $590.5$ km with a standard deviation of $s=12.5$ km.

Calculate the $p$ -value of the test and make a decision regarding $H_0$ . Round your answer to $4$ decimal places.

$p=0.0408$

On the basis of this $p$ -value, $H_0$ should be rejected, because $\,p$ $\lt$ $\alpha$ .

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.

Excel Calculation

A sample size of $n=56$ 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 $t_{n-1} = t_{55}$ distribution, under the assumption that $H_0$ is true.

Calculate the value of test statistic $t$ :
$t = \cfrac{\bar{X}-\mu_0}{s/\sqrt{n}} =\cfrac{590.5 - 587}{12.5/\sqrt{56}} = 2.09533$
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:
$=2 \text{ * }(1 \text{ - } \text{T.DIST}(\text{ABS}(t),n \text{ - } 1,1))\\ \downarrow\\ =2 \text{ * }(1 \text{ - } \text{T.DIST}(\text{ABS}(2.09533), 56 \text{ - } 1,1))$
This gives:
$p = 0.0408$
Since $\,p$ $\lt$ $\alpha$ , $H_0: \mu = 587$ should be rejected.

R Calculation

approximately has the $t_{n-1} = t_{55}$ distribution, under the assumption that $H_0$ is true.

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:
$2 \text{ * } \text{pt}(q = \text{abs}(t), df = n \text{ - } 1, lower.tail = \text{FALSE})\\ \downarrow\\ 2\text{ * } \text{pt}(q = \text{abs}(2.09533), df = 56 \text{ - } 1,lower.tail = \text{FALSE})$
This gives:
$p = 0.0408$
Since $\,p$ $\lt$ $\alpha$ , $H_0: \mu = 587$ should be rejected.

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