# 05_hypothesis_testing

## Hypothesis testing¶¶

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

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

#### Examples of formulating hypothesis¶¶

1. Nitrate levels are unsafe if > 10ppm. Test if out water is unsafe on average.

• $H_{0} => \mu \le 10ppm$
• $H_{A} => \mu > 10 ppm$
2. Test if a coin is fair.

• $H_{0} => p(h)=0.5$
• $H_{A} => p(h)\ne 0.5$ This is because alt hypothesis should not have equal sign.

#### Type 1, 2 errors¶¶

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 $\beta’ = (1-\beta)$

You calculate the test statistic as $$TS = \frac{\bar x - \mu}{\frac{s}{\sqrt n}}$$ You either reject or approve the $H_{0}$ based on the value of the TS compared against the p value for the said $\alpha$ value (confidence interval)

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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