Calculate the probability of a Type II error for the following test of hypothesis, given that $\mu = 203$.
${H_0}:\mu = 200$
${H_1}:\mu \ne 200$
$\alpha = 0.05$, $\sigma = 10$, $n = 100$
Answer
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Hint: We are given the sample mean of the test. We have to find the probability of type II error. First, we will calculate the value of z using the formula. Then, we will compare the value of z with ${z_\alpha }$. If the values differ from ${z_\alpha }$, then the null hypothesis must be rejected. Type II error is the error which is caused if we fail to reject the null hypothesis when it will come false. Then, calculate the probability of type II error by finding the area greater than value of z.
Complete step by step solution:
We are given the sample mean, $\mu = 203$ . First, we will assume that it is a two-tailed test where the value of $\alpha = 0.05$ is split into two tails. Therefore, each tail has $\dfrac{\alpha }{2}$probability. This value is called critical values.
Now, we will calculate the value of z, using the formula $z = \dfrac{{\mu - {\mu _0}}}{{\dfrac{\sigma }{{\sqrt n }}}}$
Here, $\mu = 203$, ${\mu _0} = 200$, ${z_{0.025}} = 1.96$
$ \Rightarrow z = \dfrac{{203 - 200}}{{\dfrac{{10}}{{\sqrt {100} }}}}$
On simplifying the equation, we get:
$ \Rightarrow z = \dfrac{{203 - 200}}{{\dfrac{{10}}{{10}}}}$
$ \Rightarrow z = \dfrac{{203 - 200}}{1}$
$ \Rightarrow z = 3$
Now, the value of $z = 3$ is much higher than ${z_{0.025}} = 1.96$ which means the value of the null hypothesis ${H_0}:\mu = 200$ must be rejected.
Now, the probability is calculated using the formula, $p = 2\Pr \left( {Z > 3} \right)$
$ \Rightarrow z = 3$
Now, we will compute the tail area corresponds to $Z > 3$
$ \Rightarrow 0.0013$
Since this is a two tailed test, the tail area is multiplied by two.
$ \Rightarrow 2 \times 0.0013 = 0.0026$
Thus, the probability of type II error is $0.0026$.
Note: In such types of questions students mainly forgot to convert the numbers in the same format. Students mainly make mistakes while determining whether the test is a two tailed test or one tailed test. Students may also make mistakes in deciding whether the null hypothesis is rejected or not.
Complete step by step solution:
We are given the sample mean, $\mu = 203$ . First, we will assume that it is a two-tailed test where the value of $\alpha = 0.05$ is split into two tails. Therefore, each tail has $\dfrac{\alpha }{2}$probability. This value is called critical values.
Now, we will calculate the value of z, using the formula $z = \dfrac{{\mu - {\mu _0}}}{{\dfrac{\sigma }{{\sqrt n }}}}$
Here, $\mu = 203$, ${\mu _0} = 200$, ${z_{0.025}} = 1.96$
$ \Rightarrow z = \dfrac{{203 - 200}}{{\dfrac{{10}}{{\sqrt {100} }}}}$
On simplifying the equation, we get:
$ \Rightarrow z = \dfrac{{203 - 200}}{{\dfrac{{10}}{{10}}}}$
$ \Rightarrow z = \dfrac{{203 - 200}}{1}$
$ \Rightarrow z = 3$
Now, the value of $z = 3$ is much higher than ${z_{0.025}} = 1.96$ which means the value of the null hypothesis ${H_0}:\mu = 200$ must be rejected.
Now, the probability is calculated using the formula, $p = 2\Pr \left( {Z > 3} \right)$
$ \Rightarrow z = 3$
Now, we will compute the tail area corresponds to $Z > 3$
$ \Rightarrow 0.0013$
Since this is a two tailed test, the tail area is multiplied by two.
$ \Rightarrow 2 \times 0.0013 = 0.0026$
Thus, the probability of type II error is $0.0026$.
Note: In such types of questions students mainly forgot to convert the numbers in the same format. Students mainly make mistakes while determining whether the test is a two tailed test or one tailed test. Students may also make mistakes in deciding whether the null hypothesis is rejected or not.
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