One-Sample t-Test

192. One-Sample T-Test for Mean

Population standard deviation, $𝜎$ , is unknown.

Example

An airline claims their mean departure time is 5 minutes late, but you believe their flights are more delayed than that. You take a simple random sample of 20 flights. Your sample has a mean delay of 12 mins with a sample SD of 27 mins. Assume population flight times are normally distributed. At $𝛼 = 0.05$ test the airline’s claims.

\begin{aligned} 𝐻_{0} & : 𝜇 \leq 5 \\ 𝐻_{𝑎} & : 𝜇 > 5 \end{aligned}

𝑡 = \frac{\bar{𝑥} - 𝜇}{𝑠 / \sqrt{𝑛}}

Where:

$\bar{𝑥} = 12$
$𝜇 = 5$
$𝑠 = 27$
$𝑛 = 20$
$df = 𝑛 - 1 = 19$
$𝑠 / \sqrt{𝑛} = 27 / \sqrt{20} = 6.0374$

𝑡 = \frac{12 - 5}{27 / \sqrt{20}} = 1.1594

𝑃 (𝑡 \geq 1.1594) = 0.1303

\begin{aligned} 𝑡 -value & < critical-value \\ 1.1594 & < 1.7291 \end{aligned}

\begin{aligned} 𝑝 -value & > 𝛼 -value \\ 0.1303 & > 0.05 \end{aligned}

Fail to reject $𝐻_{0}$ . At $𝛼 = 0.05$ there is not sufficient evidence that the mean departure delay is more than 5 minutes.

Code

import numpy as np
from scipy.stats import t

mu = 5
s = 27
xbar = 12
n = 20
df = n - 1
alpha = 0.05

cv = t.ppf(1 - alpha, df = df)

t_statistic = (xbar - mu) / (s / np.sqrt(n))
p = t.cdf(z)

if cv > t_statistic:
    print("Reject null hypothesis")
else:
    print("Fail to reject null hypothesis")

192.1. One-sample

Tests if the mean of a single sample differs from a known or hypothesized population mean.

𝑡 = \frac{\bar{𝑥} - 𝜇_{0}}{\frac{𝑠}{\sqrt{𝑛}}}

$\bar{𝑥}$ : sample mean
$𝜇_{0}$ : hypothesized population mean
$𝑠$ : sample standard deviation
$𝑛$ : sample size

Example

[78, 82, 89]

Step 1: State Hypotheses

𝐻_{0} : 𝜇 = 85 cm

𝐻_{1} : 𝜇 \neq 85 cm

Step 2: Summarize Data

Sample values:

𝑥_{1} = 78, 𝑥_{2} = 82, 𝑥_{3} = 89

Sample size:

𝑛 = 3

Step 3: Calculate Sample Mean ( $\bar{𝑥}$ )

\bar{𝑥} = \frac{𝑥_{1} + 𝑥_{2} + 𝑥_{3}}{𝑛} = \frac{78 + 82 + 89}{3} = 83

Step 4: Calculate Sample Standard Deviation:

𝑠 = \sqrt{\frac{\sum_{𝑖 = 1}^{𝑛} {(𝑥_{𝑖} - \bar{𝑥})}^{2}}{𝑛 - 1}}

Find the deviations from the mean and square them

{(𝑥_{1} - \bar{𝑥})}^{2} = {(78 - 83)}^{2} = {(- 5)}^{2} = 25

{(𝑥_{2} - \bar{𝑥})}^{2} = {(82 - 83)}^{2} = {(- 1)}^{2} = 1

{(𝑥_{3} - \bar{𝑥})}^{2} = {(89 - 83)}^{2} = {(6)}^{2} = 36

Sum of squared deviations

SSD = 25 + 1 + 36

Calculate variance of sample

𝑠^{2} = \frac{SSD}{𝑛 - 1} = \frac{62}{3 - 1} = \frac{62}{2} = 31

Calculate Sample Standard Deviation

𝑠 = \sqrt{𝑠^{2}} = \sqrt{32} = 5.57

Step 5: Calculate the Test Statistic

𝑡 = \frac{\bar{𝑥} - 𝜇_{0}}{\frac{𝑠}{\sqrt{𝑛}}} = \frac{83 - 85}{\frac{5.57}{\sqrt{3}}} = - \frac{3}{3.22} = - 0.62

Step 6: Determine the Degrees of Freedom

𝑑 𝑓 = 𝑛 - 1 = 15 - 1 = 14

Step 7: Find Critical t-value

For a two-tailed test at a significance level ( $𝛼$ ) of 0.05 and 2 degrees of freedom ( $𝑑 𝑓$ )

4.303

Step 8: Compare the t-Value to the Critical t-Value

If the absolute value of the test statistic is greater than the critical t-value, reject the null hypothesis.
If the absolute value of the test statistic is less than the critical t-value, fail to reject the null hypothesis.

Step 8: Find the p-Value

𝑡 = 4.303 and 𝑑 𝑓 = 3

p-value = 0.58

t_test_one_sample.py

from scipy import stats
rvs = stats.uniform.rvs(size=50)
stats.ttest_1samp(rvs, popmean=0.5)