The goal of hypothesis testing is to answer a simple **yes / no** question about a population parameter. There are two types of hypothesis, $H_{0}$ the **null hypothesis** and $H_{A}$ the **Alternate hypothesis**.

The steps followed are:

- set up the hypothesis (null, alternate)
- choose $\alpha$ level (confidence interval)
- determine rejection region (on the z curve)
- compute the test statistic (p value based on z score)
- make a decision

**Rules in hypothesis testing**

- No
*equal*sign in $H_{A}$. Only $\ne, <, >$ signs - Put what you want to test in $H_{A}$, unless you violate rule 1, then you put that in $H_{0}$
- Believe $H_{0}$ unless the dataset shows otherwise
- when we make our decision, we either reject $H_{0}$ or fail to reject it.

For a jury trial, our motto is **innocent until proven guilty**. Hence

- $H_{0} => innocent$ as we reject H0 or fail to do so
$H_{A} => guilty$

Type 1 error:

**False positive**- we reject $H_{0}$ when it is still true
- $\alpha$ = p(type 1 error) = p(rejecting $H_{0}$ when it is still true)

- Type 2 error:
**False negative**- $\beta$ = p(type 2 error) = p(failing to reject $H_{0}$ when $H_{A}$ is true)

In practice, we fix $\alpha = 0.5$ and calculate $\beta' = (1-\beta)$

**Example**
A sample of `49`

batteries are tested for their limetimes. The SD is `15.0`

, mean longivity is `1006.2`

. Is it possible to claim the batteries last longer than `1000`

hours on average?

$\bar x = 1006.2$, $n=49$, $s=15$, $\alpha = 0.01$ assumed. $$H_{0} => \mu \le 1000$$ $$H_{A} => \mu > 1000$$

Find Test Statistic $$TS = \frac{1006.2-1000}{15/\sqrt{49}}$$ $$TS=2.89$$ This is a right tailed hypothesis as we test if test statistic is > z score for the said alpha.

z score for $\alpha=0.01$ = 2.576 (for 99% CI)
The TS is > z score. Hence **reject $H_{0}$. Thus mean battery life > 1000 hours** by significance.

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