# t-test calculator for Mean with Examples

## One Sample $t$-Test for mean

In this tutorial we will discuss $t$-test calculator for testing population mean along with step by step numerical examples on $t$-test for testing single population mean.

## One Sample $t$-test Calculator for mean

The one sample $t$-test for mean is used to perform hypothesis testing problem for population mean when the population standard deviation is unknown. This calculator makes it easy to conduct one sample $t$-test for population mean.

t-test Calculator for mean
Population Mean ($\mu$)
Sample Size ($n$)
Sample Mean ($\overline{x}$)
Sample Standard Deviation ($s$)
Level of Significance ($\alpha$)
Tail: Left tailed
Right tailed
Two tailed
Results
Standard Error of Mean:
Test Statistics t:
Degrees of Freedom:
t-critical value:
p-value:

## How to use $t$-test calculator for testing single proportion?

Step 1 - Enter the population mean $\mu$

Step 2 - Enter the sample size $n$

Step 3 - Enter the sample mean $\overline{X}$ and sample standard deviation

Step 4 - Enter the level of significance $\alpha$

Step 5 - Select the alternative hypothesis (left-tailed / right-tailed / two-tailed)

Step 6 - Click on "Calculate" button to get the result

## $t$-test for mean Example 1

The Healthy Food Company claims that its cereal boxes contain, on average, 453 grams of cereal. We suspect that the cereal boxes contain, on average, less than claimed. You decided to test the claim by inspecting 6 randomly selected boxes, and get the following weights:

454, 447, 452, 446, 450, 445.

Assume that the amount of cereal in a box follows a normal distribution. Test at 5% level of significance.

#### Solution

Given that the sample size $n = 6$, sample mean $\overline{x}= 449$ and sample standard deviation $s = 3.578$.

#### Step 1 Hypothesis Testing Problem

The hypothesis testing problem is
$H_0 : \mu = 453$ against $H_1 : \mu < 453$ ($\text{left-tailed}$)

#### Step 2 Test Statistic

The test statistic is

 \begin{aligned} t& =\frac{\overline{x} -\mu}{s/\sqrt{n}} \end{aligned}
which follows $t$ distribution with $n-1$ degrees of freedom.

#### Step 3 Significance Level

The significance level is $\alpha = 0.05$.

#### Step 4 Critical Value(s)

As the alternative hypothesis is $\text{left-tailed}$, the critical value of $t$ $\text{is}$ $-2.015$.

The rejection region (i.e. critical region) is $\text{t < -2.015}$.

#### Step 5 Computation

The test statistic under the null hypothesis is

 \begin{aligned} t&=\frac{ \overline{x} -\mu_0}{s/\sqrt{n}}\\ &= \frac{449-453}{3.578/ \sqrt{6 }}\\ &= -2.738 \end{aligned}

#### Step 6 Decision (Traditional Approach)

The test statistic is $t =-2.738$ which falls $\text{inside}$ the critical region, we $\text{reject}$ the null hypothesis.

OR

#### Step 6 Decision ($p$-value Approach)

This is a $\text{left-tailed}$ test, so the p-value is the area to the left of the test statistic ($t=-2.738$) is p-value = $0.0204$.

The p-value is $0.0204$ which is $\text{less than}$ the significance level of $\alpha = 0.05$, we $\text{reject}$ the null hypothesis.

## $t$-test for mean Example 2

A new brand of laptop battery is produced by a company. The company claims that the battery will last for an extended period of time before a recharge is necessary. A sample of 40 batteries is tested for the length of usage time to recharge. The sample results are as follows:

Sample Size = 40
Sample Mean = 6.5 hrs.
Sample Standard Deviation = 1.3 hrs.

At the 0.01 level of significance, is there evidence that the population mean length of usage is greater than 5.5 hours?

#### Solution

Given that the sample size $n = 40$, sample mean $\overline{x}= 6.5$ and sample standard deviation $s = 1.3$.

#### Step 1 Hypothesis Testing Problem

The hypothesis testing problem is
$H_0 : \mu = 5.5$ against $H_1 : \mu > 5.5$ ($\text{right-tailed}$)

#### Step 2 Test Statistic

The test statistic is

 \begin{aligned} t& =\frac{\overline{x} -\mu}{s/\sqrt{n}} \end{aligned}
which follows $t$ distribution with $n-1$ degrees of freedom.

#### Step 3 Significance Level

The significance level is $\alpha = 0.01$.

#### Step 4 Critical Value(s)

As the alternative hypothesis is $\text{right-tailed}$, the critical value of $t$ $\text{is}$ $2.426$.

The rejection region (i.e. critical region) is $\text{t > 2.426}$.

#### Step 5 Computation

The test statistic under the null hypothesis is

 \begin{aligned} t&=\frac{ \overline{x} -\mu_0}{s/\sqrt{n}}\\ &= \frac{6.5-5.5}{1.3/ \sqrt{40 }}\\ &= 4.865 \end{aligned}

#### Step 6 Decision (Traditional Approach)

The test statistic is $t =4.865$ which falls $\text{inside}$ the critical region, we $\text{reject}$ the null hypothesis.

OR

#### Step 6 Decision ($p$-value Approach)

This is a $\text{right-tailed}$ test, so the p-value is the area to the left of the test statistic ($t=4.865$) is p-value = $0$.

The p-value is $0$ which is $\text{less than}$ the significance level of $\alpha = 0.01$, we $\text{reject}$ the null hypothesis.

## $t$-test for mean Example 3

A special cable manufactured by a company has a breaking strength of 800 pounds. (Assume the breaking strength is normally distributed.) A researcher selects a random sample of 20 cables and finds the average breaking strength is 793 pounds with a standard deviation of 12 pounds. Do these data support the claim that the average breaking strength is not 800 pounds? Test at the 0.05 level of significance.

#### Solution

Given that the sample size $n = 20$, sample mean $\overline{x}= 793$ and sample standard deviation $s = 12$.

#### Step 1 Hypothesis Testing Problem

The hypothesis testing problem is
$H_0 : \mu = 800$ against $H_1 : \mu \neq 800$ ($\text{two-tailed}$)

#### Step 2 Test Statistic

The test statistic is

 \begin{aligned} t& =\frac{\overline{x} -\mu}{s/\sqrt{n}} \end{aligned}
which follows $t$ distribution with $n-1$ degrees of freedom.

#### Step 3 Significance Level

The significance level is $\alpha = 0.05$.

#### Step 4 Critical Value(s)

As the alternative hypothesis is $\text{two-tailed}$, the critical value of $t$ $\text{are}$ $-2.093 and 2.093$.

The rejection region (i.e. critical region) is $\text{t < -2.093 or t > 2.093}$.

#### Step 5 Computation

The test statistic under the null hypothesis is

 \begin{aligned} t&=\frac{ \overline{x} -\mu_0}{s/\sqrt{n}}\\ &= \frac{793-800}{12/ \sqrt{20 }}\\ &= -2.609 \end{aligned}

#### Step 6 Decision (Traditional Approach)

The test statistic is $t =-2.609$ which falls $\text{inside}$ the critical region, we $\text{reject}$ the null hypothesis.

OR

#### Step 6 Decision ($p$-value Approach)

This is a $\text{two-tailed}$ test, so the p-value is the area to the left of the test statistic ($t=-2.609$) is p-value = $0.0173$.

The p-value is $0.0173$ which is $\text{less than}$ the significance level of $\alpha = 0.05$, we $\text{reject}$ the null hypothesis.

## Endnote

In this tutorial, you learned the about how to solve numerical examples on $t$-test for testing population mean. You also learned about the step by step procedure to apply $t$-test for testing population mean and how to use $t$-test calculator for testing population mean to get the value of test statistic, p-value, and t-critical value.

Let me know in the comments if you have any questions on $t$-test calculator for mean with examples and your thought on this article.