Hypotheses Testing

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Hypotheses Testing. Example 1. We have tossed a coin 50 times and we got k = 19 heads Should we accept/reject the hypothesis that p = 0.5 (the coin is fair). Null versus Alternative. Null hypothesis (H 0 ): p = 0.5 - PowerPoint PPT Presentation

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Hypotheses Testing

Example 1

We have tossed a coin 50 times and we got

k = 19 heads

Should we accept/reject the hypothesis that p = 0.5(the coin is fair)

Null versus Alternative

Null hypothesis (H0): p = 0.5

Alternative hypothesis (H1): p 0.5

0 5 10 15 20 25 30 35 40 45 500

0.02

0.04

0.06

0.08

0.1

0.12

k

p(k)

95%

EXPERIMENT

Significance level α= Probability of Type 1 error

=Pr[rejecting H0 | H0 true]

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

If k < 18 or k > 32 then under the null hypothesis the observed event falls into the rejection region with probability α < 0.05

We want α as small as possible

Test construction

18 32

acceptreject reject

0 5 10 15 20 25 30 35 40 45 500

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0.025

0.975

k

Cpdf(k)

Conclusion

No evidence to reject the null hypothesis

Example 2

We have tossed a coin 50 times and we got

k = 10 heads

Should we accept/reject the hypothesis that p = 0.5(the coin is fair)

0 5 10 15 20 25 30 35 40 45 500

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

k

cpdf(k)

p-value P[ k 10 or k 40 ] 0.000025

We could REJECT hypothesis H0

at significance level as low as α= 0.000025

p-value is the lowest attainable sig level

Remark

In STATISTICS

To prove something = REJECT the hypothesis that converse is true

Example 3

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 our group and we get sample = 5.5, 5.6, 4.3, 5.1, 5.2, 6.1, 5.0, 5.2, 5.8, 4.1

Hypothesis H0 - sample = sample from normal distribution with = 5cm

Alternative H1 - sample = sample from normal distribution with > 5cm

Construction of the test

tt0.95

reject

Cannot reject

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

distribution

N

iiXN

X1

1

N

ii XX

NS

1

21

1

NS

Xt

2 test (K. Pearson, 1900)

To test the hypothesis that a given data actually come from a population with the proposed distribution

Data 0.4319 0.6874 0.5301 0.8774 0.6698 1.1900 0.4360 0.2192 0.5082 0.3564 1.2521 0.7744 0.1954 0.3075 0.6193 0.4527 0.1843 2.2617 0.4048 2.3923 0.7029 0.9500 0.1074 3.3593 0.2112 0.0237 0.0080 0.1897 0.6592 0.5572 1.2336 0.3527 0.9115 0.0326 0.2555 0.7095 0.2360 1.0536 0.6569 0.0552 0.3046 1.2388 0.1402 0.3712 1.6093 1.2595 0.3991 0.3698 0.7944 0.4425 0.6363 2.5008 2.8841 0.9300 3.4827 0.7658 0.3049 1.9015 2.6742 0.3923 0.3974 3.3202 3.2906 1.3283 0.4263 2.2836 0.8007 0.3678 0.2654 0.2938 1.9808 0.6311 0.6535 0.8325 1.4987 0.3137 0.2862 0.2545 0.5899 0.4713 1.6893 0.6375 0.2674 0.0907 1.0383 1.0939 0.1155 1.1676 0.1737 0.0769 1.1692 1.1440 2.4005 2.0369 0.3560 1.3249 0.1358 1.3994 1.4138 0.0046

Are these data sampled from population with exponential pdf ?

xexf )(

Construction of the 2 test

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

p1p2 p3 p4

Construction of the test

2

2 0.95

reject

Cannot reject

How aboutAre these data sampled from population with exponential pdf ?

axaexf )(

1. Estimate a2. Use 2 test3. Remember d.f. = K-2

Power and significance of the test

Actual situation

decision probability

H0 true

H0 false

accept

Reject = error t. I

reject

Accept = error t. II

1-α

α = significance level

β

1-β = power of the test