Type I Error

The random variable \(X\) is binomially distributed. Suppose a sample of 10 is taken and a statistician wants to test the null hypothesis \(H_0: \; p=0.45\) against the alternative hypothesis \(H_1:\; p\neq0.45\).

a) Find the critical region for this test.

b) State the probability of a Type I error for this test.

Solution:

a) Since this is a two tailed test, at a \(5\%\) significance level, the critical values, \(c_1\) and \(c_2\) are such that

\[\begin{align} \mathbb{P}(X\leq c_1) &\leq0.025 \\ \text{ and } \mathbb{P}(X\geq c_2) &\leq 0.025. \end{align}\]

\(\mathbb{P}(X\geq c_2) = 1-\mathbb{P}(X\leq c_2-1)\leq0.025\) or \( \mathbb{P}(X\leq c_2-1) \geq0.975\)

Assume \(H_0\) is true. Then under the null-hypothesis \(X\sim B(10,0.45)\), from the statistical tables:

\[ \begin{align} &\mathbb{P}(X \leq 1)=0.0233<0.025 \\ & \mathbb{P}(X \leq 2)=0.0996>0.025.\end{align}\]

Therefore the critical value is \(c_1=1\). For the second critical value,

\[ \begin{align} &\mathbb{P}(X \leq 7)=0.9726<0.975 \\ & \mathbb{P}(X \leq 8)=0.996>0.975. \end{align}\]

Therefore \(c_2-1=8\) so the critical value is \(c_2=9\).

So the critical region for this test under a \(5\%\) significance level is

\[\left\{ X\leq 1\right\}\cup \left\{ X\geq 9\right\}.\]

b) A Type I error occurs when you reject \(H_0\) but \(H_0\) is true, i.e. it is the probability you are in the critical region given that the null hypothesis is true.

Under the null hypothesis, \(p=0.45\), therefore,

\[\begin{align} \mathbb{P}(\text{Type I error})&=\mathbb{P}(X\leq1 \mid p=0.45)+\mathbb{P}(X\geq9 \mid p=0.45) \\ &=0.0233+1-0.996 \\ &=0.0273. \end{align}\]

A coin is tossed until a tail is obtained.

a) Using a suitable distribution, find the critical region for a hypothesis test that tests whether the coin is biased towards heads at the \(5\%\) significance level.

b) State the probability of a Type I error for this test.

Solution:

a) Let \(X\) be the number of coin tosses before a tail is obtained.

Then this can be answered using the geometric distribution as follows since the number of failures (heads) \(k - 1\) before the first success/tail with a probability of a tail given by \(p\).

Therefore, \(X\sim \rm{Geo}(p)\) where \(p\) is the probability of a tail being obtained. Therefore the null and alternative hypothesis are

\[ \begin{align} &H_0: \; p=\frac{1}{2} \\ \text{and } &H_1: \; p<\frac{1}{2}. \end{align}\]

Here the alternative hypothesis is the one that you want to establish, i.e. that the coin is biased towards heads, and the null hypothesis is the negation of that, i.e. the coin is not biased.

Under the null hypothesis \(X\sim \rm{Geo} \left(\frac{1}{2}\right)\).

Since you are dealing with a one-tailed test at the \(5\%\) significance level, you want to find the critical value \(c\) such that \(\mathbb{P}(X\geq c) \leq 0.05 \). This means you want

\[ \left(\frac{1}{2}\right)^{c-1} \leq 0.05. \]

Therefore

\[ (c-1)\ln\left(\frac{1}{2}\right) \leq \ln(0.05), \]

which means \(c >5.3219\).

Therefore, the critical region for this test is \(X \geq 5.3219=6\).

Here you have used the fact that, for a geometric distribution \(X\sim \rm{Geo}(p)\),

\[\mathbb{P}(X \geq x)=(1-p)^{x-1}.\]

b) Since \(X\) is a discrete random variable, \(\mathbb{P}(\text{Type I error})\leq \alpha\), and the probability of a Type I error is the actual significance level. So

\[\begin{align} \mathbb{P}(\text{Type I error})&= \mathbb{P}( \text{rejecting } H_0 \text{ when } H_0 \text{ is true}) \\ &=\mathbb{P}(X\geq 6 \mid p=0.5) \\ &= \left(\frac{1}{2}\right)^{6-1} \\ &=0.03125. \end{align}\]

The random variable \(X\) is normally distributed such that \(X\sim N(\mu ,4)\). Suppose a random sample of \(16\) observations is taken and \(\bar{X}\) the test statistic. A statistician wants to test \(H_0:\mu=30\) against \(H_1:\mu<30\) using a \(5\%\) significance level.

a) Find the critical region.

b) State the probability of a Type I error.

Solution:

a) Under the null hypothesis you have \(\bar{X}\sim N(30,\frac{4}{16})\).

Define

\[Z=\frac{\bar{X}-\mu}{\frac{\mu}{\sqrt{n}}}\sim N(0,1).\]

At the \(5\%\) significance level for a one-sided test, from the statistical tables, the critical region for \(Z\) is \(Z<-1.6449\).

Therefore, you reject \(H_0\) if

\[\begin{align} \frac{\bar{X}-\mu}{\frac{\mu}{\sqrt{n}}}&=\frac{\bar{X}-30}{\frac{2}{\sqrt{16}}} \\ &\leq -1.6449.\end{align}\]

Therefore, with some rearranging, the critical region for \(\bar{X}\) is given by \(\bar{X} \leq 29.1776\).

b) Since \(X\) is a continuous random variable, there is no difference between the target significance level and the actual significance level. Therefore, \(\mathbb{P}(\text{Type I error})= \alpha\) i.e. the probability of a Type I error \(\alpha\) is the same as the significance level of the test, so

\[\mathbb{P}(\text{Type I error})=0.05.\]