4.4 Hypotheses testing

We have tossed a coin 50 times and we got k = 19 heads . Should we accept/reject the hypothesis that p = 0.5, provided taht the coin is fair?

Null versus alternative hypothesis:

• Null hypothesis $\left(H_0\right):p=\mathrm{0.5.}$
• Alternative hypothesis $\left(H_1\right):p\ne \mathrm{0.5.}$

Significance level $\alpha$ = Probability of Type I error = Pr[rejecting $H_0$ | $H_0$ true]

P [ $k<18$ or $k>32$ ] $<0.05$ .

If $k<18$ or $k>32$ ] $<0.05$ , then under the null hypothesis the observed event falls into rejection region with the probability $\alpha <\mathrm{0.05.}$

We have tossed a coin 50 times and we got k = 10 heads . Should we accept/reject the hypothesis that p = 0.5, provided taht the coin is fair?

P [ $k\le 10$ or $k\ge 40$ ] $\approx \mathrm{0.000025.}$ We could reject hypothesis $H_0$ at a significance level as low as $\alpha =\mathrm{0.000025.}$

We know that on average mouse tail is 5 cm long. We have a group of 10 mice, and give to each of them a dose of vitamin T everyday, from the birth, for the period of 6 months.

We want to prove that vitamin X makes mouse tail longer. We measure tail lengths of out group and we get the following sample:

• Hypothesis $H_0$ - sample = sample from normal distribution with $\mu$ = 5 cm.
• Alternative $H_1$ - sample = sample from normal distribution with $\mu$ >5 cm.

We do not know population variance, and/or we suspect that vitamin treatment may change the variance - so we use t distribution .

• $\overline{X}=\frac{1}{N}{\displaystyle \sum _{i=1}^N{X}_i},$
• $S=\sqrt{\frac{1}{N}{\displaystyle \sum _{i=1}^N{\left(X_i-\overline{X}\right)}^2}},$
• $t=\frac{\overline{X}-\mu }{S}\sqrt{N-1}.$

${\chi }^2$ test (K. Pearson, 1900)

To test the hypothesis that a given data actually come from a population with the proposed distribution. Data is given in the Table 2 .