This tutorial explains how to use Excel and SPSS to perform one sample T Test.

You may also want to read:

Calculate Z Score and probability using SPSS and Excel

## What is T Test

In statistical inference, we are interested to know whether a small sample comes from a population. To inference using sample mean, when the population standard deviation and population mean are known, we can use Z test to interference the population mean from sample mean.

In reality, we do not have data of the whole population. Without the population standard deviation, we use T Test (also known as **Student’s t Test**) to interference the population mean from sample mean.

There are basically three kinds of T Test in statistics:

One sample T Test | Test whether a target mean value is equal to the population mean |

Independent samples T Test | Test whether the population mean from two individual samples are equal |

Paired-samples T Test | Test whether the population mean between paired observations are equal |

This tutorial will only talk about One sample T Test.

## One sample T Test – manual calculation

### Formula

t value of particular test mean value

x bar: sample mean

μo: population mean to test

s: sample standard deviation

n: sample size

Confidence interval estimator of population mean (the value range of population mean at particular confidence level α)

Degree of freedom (ν)

ν = n-1

### Example

To test whether population mean of student’s exam marks is over 50, given the below sample data

Student ID | Average marks |

1 | 40 |

2 | 50 |

3 | 60 |

4 | 36 |

5 | 70 |

#### Step 1: Hypothesis

H1: μ > 50

Ho: μ = 50

Click here to learn more about Hypothesis Testing

#### Step 2: Calculate t of test mean value

sample mean = 51.2

sample standard deviation = 14.043

square root of sample size = 2.236

t = (51.2-50)/(14.043/2.236)

t = 0.191

#### Step 3: Find t at 5% Significance level from t table

Find the t value for df = 4, p = 0.05 in t table

The value of t is 2.132

#### Conclusion

The rejection region is t > 2.132. Now that t for 50 is 0.191, therefore we do not have enough evidence to reject null hypothesis.

We can also calculate the confidence interval for our reference.

Interval = 2.571*14.043/2.236 = 16.15

Thus

Lower limit = 50-16.15 = 33.85

Upper limit = 50+16.15 = 66.15

## One sample T Test – Excel

Create a spreadsheet as below. Column A is optional, column C is a dummy column to be used later as explained below.

In Excel 2010 / 2013, navigate to Data > Data Analysis

Since Excel does not have One Sample T Test, the closet we can use is **t-Test: Two Sample Assuming Unequal Variances. **We will compare column B with column C as a workaround.

Enter the information as below, and the click OK.

**Variable 1 Range** is your data source.** Variable 2 Range** is the hypothesis mean in this case.

**Hypotheses Mean Difference** is 0 because we just want to know if there is any difference, but not how much difference.

**Labels** means whether the first data of variable range is a header, which will be used in the output table.

**Alpha** is the significance level of one tail.

**Output Range** is where we want to output the result table.

This is the generated result table. I have highlighted the two T value for easy reference. They are same as manual calculation.

To find the confidence interval in Excel, navigate to Data > Data Analysis

Select Descriptive Statistics

Input the value as below, click on OK

A table is generated. The value I highlight is the confidence interval, therefore

Lower limit = 50-17.44 = 32.56

Upper limit = 50+17.44 = 67.44

## One sample T Test – SPSS

Navigate to Analyze > Compare Means > One-Sample T Test

Move the average marks field to the right box for analysis, and then enter 50 as Test Value, click on OK.

#### Interpreting the result:

t | This is the t value for mean = 50 |

df | degree of freedom equals to sample size-1 |

Sig (2-tailed) | The total probability that mean > 50 and mean < 50 is 0.858 (probability not equal to 50). Since our hypothesis is testing mean > 50, we divide 0.858 by 2 = 0.429. Since 0.429 > 0.05 significance level, we cannot reject Null hypothesis. |

Mean Difference | Difference between sample mean (51.2) and test value (50) |

95% Confidence Interval of the Difference | As we do not know the true mean of the population, from here we can estimate 95% chance that the mean would lie between 34.8 (51.2-16.24) and 69.84 (51.2+18.64). |

95% confidence level is the default value, to change the confidence level, go back to the **One-Sample T Test** box > click on **Options** button

You may change the confidence level right here

## Outbound References

https://www.youtube.com/watch?v=tdaVp9BOdjk