{"id":3471,"date":"2026-03-01T20:31:39","date_gmt":"2026-03-01T19:31:39","guid":{"rendered":"https:\/\/www.gironi.it\/blog\/?p=3471"},"modified":"2026-07-17T08:47:25","modified_gmt":"2026-07-17T07:47:25","slug":"the-t-distribution-and-hypothesis-testing","status":"publish","type":"post","link":"https:\/\/www.gironi.it\/blog\/en\/the-t-distribution-and-hypothesis-testing\/","title":{"rendered":"T-Test with R examples for SEO: comparing CTR, conversions and metrics"},"content":{"rendered":"\n

In the article on hypothesis testing<\/a> we saw how to check whether a sample mean is compatible with a hypothesis about the population. In the examples we proposed, however, we knew the population standard deviation, sigma<\/strong> \u2014 which allowed us to use the normal distribution<\/strong> and the Z-score<\/strong>.<\/p>\n\n\n\n

In practice this is a rather rare case. If we do not know the population sigma, or if we are working with small samples<\/strong>, we need a different distribution: the t distribution<\/strong>, or Student’s distribution<\/strong>.<\/p>\n\n\n\n\n\n\n\n

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Student’s t distribution is a probability distribution used to assess the statistical significance of results when the sample is small and the population variance is unknown.<\/strong><\/p>\n<\/div><\/div>\n\n\n\n

A Brief Historical Digression<\/h2>\n\n\n\n
\"William
William Sealy Gosset (Student)<\/figcaption><\/figure>\n\n\n\n

In the early 1900s, the chemist and statistician William Sealy Gosset, employed at the Guinness brewery (and a collaborator of Karl Pearson), discovered that when working with very small samples, the distributions of the mean differed significantly from the normal. As the sample size changed, the shape of the distribution changed too \u2014 and as the sample grew, it gradually approximated the normal.<\/p>\n\n\n\n

Unable to reveal his identity so as not to benefit competitors, he published under the pseudonym “Student”. This is why distributions for small samples are now known as Student’s t distributions<\/strong>.<\/p>\n\n\n\n

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The t distribution is symmetric about zero, but it is “flatter” than the standardised normal: a greater portion of its area lies in the tails.<\/p>\n\n\n\n

A larger sample causes the t distribution to approximate the normal ever more closely. The differences are greatest when we have few degrees of freedom.<\/strong><\/p>\n\n\n\n

\"Student's
Student’s t distributions for several degrees of freedom, compared with the normal (dashed). With few degrees of freedom the tails are heavier; by df = 30 the t is almost indistinguishable from the normal.<\/figcaption><\/figure>\n\n\n\n

What do we mean by degrees of freedom<\/strong>? The number of sample values that have the “freedom” to vary without changing the sample mean. In practice, for a one-sample test:<\/p>\n\n\n\n\\( df = n – 1 \\\\ \\)\n\n\n\n

where n is the sample size.<\/p>\n\n\n\n

The procedure for hypothesis testing with the t distribution largely mirrors what we saw with the normal. We state the null hypothesis H\u2080 and the alternative hypothesis H\u2090<\/strong>, then compute the test statistic:<\/p>\n\n\n\n\\( t = \\frac{\\bar{x} – \\mu}{\\frac{s}{\\sqrt{n}}} \\\\ \\)\n\n\n\n

where \\( \\frac{s}{\\sqrt{n}} \\) is the estimated standard error<\/strong> (also denoted SE).<\/p>\n\n\n\n

An Example Is Worth a Thousand Explanations<\/h2>\n\n\n\n

A light bulb manufacturer claims that its product has a mean lifespan of at least 4200 hours. We take a sample of n = 10 bulbs and find a sample mean of 4000 hours, with a sample standard deviation of 200 hours.<\/p>\n\n\n\n\\( n = 10 \\quad \\bar{x} = 4000 \\quad s = 200 \\\\ \\)\n\n\n\n

We set up the test:<\/p>\n\n\n\n

\\( H_0 : \\mu \\ge 4200 \\qquad H_a : \\mu < 4200 \\)<\/p>\n\n\n\n

We choose a significance level \u03b1 = 0.05<\/strong> (corresponding to a 95% confidence level). In the t distribution critical value table, we look up the value for 9 degrees of freedom and \u03b1 = 0.05 (one-tailed). The critical value is 1.833.<\/strong> We will reject H\u2080 if the computed t is less than \u22121.833.<\/p>\n\n\n\n

The standard error:<\/p>\n\n\n\n

\\[ SE = \\frac{s}{\\sqrt{n}} = \\frac{200}{\\sqrt{10}} = 63.3 \\]<\/p>\n\n\n\n

And the t value:<\/p>\n\n\n\n

\\[ t = \\frac{\\bar{x} – \\mu}{SE} = \\frac{4000 – 4200}{63.3} = -3.16 \\]<\/p>\n\n\n\n

The t value falls in the critical region<\/strong>: we reject H\u2080. There is sufficient statistical evidence to conclude that the mean lifespan is less than the claimed 4200 hours.<\/p>\n\n\n\n

\"One-sample
One-sample t-test: critical region and observed value<\/figcaption><\/figure>\n\n\n\n

The p-Value<\/h2>\n\n\n\n

We can also evaluate the hypothesis by asking: what is the probability of obtaining a test statistic value equal to or more extreme than the one observed, assuming H\u2080 is true? This probability is the p-value<\/strong>.<\/p>\n\n\n\n

Using R or a calculator we obtain t = \u22123.16 and p = 0.00578. The commands for the calculators:<\/p>\n\n\n\n