# Suppose X ? N ( ? , 36 ) . From a sample of size n = 16, we wish to test H 0 : ? = 5 v s . H...

## Question:

Suppose {eq}X \sim N(\mu , 36) {/eq} . From a sample of size n = 16, we wish to test {eq}H_0: \mu = 5 vs. H_1: \mu > 5 {/eq}

a. What is the rejection region for a test of size .05?

b. What is the power of your test if mu = 6?

## Power of a test

Power of the test defines the probability that the null hypothesis would be rejected correctly. This infers that alternate hypothesis was true and hence the null is correctly rejected. It is calculated as the complement of the type II error rate, which defines the probability of accepting the null hypothesis given it is false.

## Answer and Explanation:

**Given Information**

{eq}X \sim N\left( {\mu ,36} \right) {/eq}

Sample size is 16 (n)

Hypotheses is given as,

{eq}\begin{align*} {H_o}:\mu = 5\\ {{\rm H}_1}:\mu > 5 \end{align*} {/eq}

**(a)**

Significance level is {eq}0.05\left( \alpha \right) {/eq}

At the significance level, the z-value that defines the right tailed rejection region is obtained from Z-table,

{eq}\begin{align*} P\left( {Z > {\rm{1}}{\rm{.645}}} \right) &= 1 - P\left( {Z < {\rm{1}}{\rm{.645}}} \right)\\ &= 1 - 0.95\\ &= 0.05 \end{align*} {/eq}

Hence the rejection region in terms of Z-scores is defined as

{eq}\left[ {Z:Z > 1.645} \right] {/eq}

Now the corresponding critical value is calculated using the formula,

{eq}\begin{align*} {Z_{crit}} &= \dfrac{{{X_{crit}} - \mu }}{{\dfrac{\sigma }{{\sqrt n }}}}\\ {X_{crit}} &= \left( {1.645 \times \dfrac{6}{{\sqrt {16} }}} \right) + 5\\ &= 7.467 \end{align*} {/eq}

**(b)**

The power of a test {eq}P = 1 - \beta {/eq}, where \beta is type II error, which is probability of accepting the null hypothesis, given the true mean value is different from hypothesized means value.

{eq}\begin{align*} \beta &= P\left( {X < {X_{crit}}\;{\rm{given}}\;\mu = 6} \right)\\ &= P\left( {Z < \dfrac{{{X_{crit}} - \mu }}{{\dfrac{\sigma }{{\sqrt n }}}}} \right)\\ &= P\left( {Z < \dfrac{{7.467 - 6}}{{\dfrac{6}{{\sqrt {16} }}}}} \right)\\ & = P\left( {Z < 0.978} \right) \end{align*} {/eq}

Calculating probability using Z-table

{eq}\begin{align*} \beta &= {\rm{0}}{\rm{.836}}\\ P &= 1 - 0.836\\ &= 0.164 \end{align*} {/eq}

Hence power of the test is **0.164**

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from Statistics 101: Principles of Statistics

Chapter 10 / Lesson 4