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HYPOTHESES TESTING

Module by: Ewa Paszek. E-mail the author

Summary: This course is a short series of lectures on Introductory Statistics. Topics covered are listed in the Table of Contents. The notes were prepared by Ewa Paszek and Marek Kimmel. The development of this course has been supported by NSF 0203396 grant.

Hypotheses Testing - Examples.

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, provided taht the coin is fair?

Null versus Alternative Hypothesis:

  • Null hypothesis ( H 0 ):p=0.5. ( H 0 ):p=0.5.
  • Alternative hypothesis ( H 1 ):p≠0.5. ( H 1 ):p≠0.5.

Figure 1:
EXPERIMENT
EXPERIMENT (p1.gif)

Significance level α α = Probability of Type I error = Pr[rejecting H 0 H 0 | H 0 H 0 true]

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

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

Note that:

We want α α as small as possible.
Figure 2
(a) Test construction.
Figure 2(a) (Slide000.gif)
(b) Cumulative distribution function.
Figure 2(b) (p2.gif)

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, provided taht the coin is fair?

Figure 3: Cumulative distribution function.
EXPERIMENT
EXPERIMENT (p2.gif)

P[ k≤10 k≤10 or k≥40 k≥40 ] ≈0.000025. ≈0.000025. We could reject hypothesis H 0 H 0 at a significance level as low as α=0.000025. α=0.000025.

Note That:
p-value is the lowest attainable significance 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 out group and we get the following sample:

Table 1: Table 1
5.5 5.6 4.3 5.1 5.2 6.1 5.0 5.2 5.8 4.1
  • Hypothesis H 0 H 0 - sample = sample from normal distribution with μ μ = 5 cm.
  • Alternative H 1 H 1 - sample = sample from normal distribution with μ μ > 5 cm.
Figure 4:
CONSTRUCTION OF THE TEST
CONSTRUCTION OF THE TEST (Slide155.gif)

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

  • X ¯ = 1 N ∑ i=1 N X i , X ¯ = 1 N ∑ i=1 N X i ,
  • S= 1 N ∑ i=1 N ( X i − X ¯ ) 2 , S= 1 N ∑ i=1 N ( X i − X ¯ ) 2 ,
  • t= X ¯ −μ S N−1 . t= X ¯ −μ S N−1 .

Example 4

χ 2 χ 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.

Table 2: 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 - - - - - - - -
Problem 1

Are these data sampled from population with exponential p.d.f.?

Solution

f( x )= e −x . f( x )= e −x .

Figure 5
CONSTRUCTION OF THE TEST
(a)
Figure 5(a) (Slide002.gif)
(b)
Figure 5(b) (Slide001.gif)

Exercise 1

Are these data sampled from population with exponential p.d.f.?

Solution

f( x )=a e −ax . f( x )=a e −ax .

  1. Estimate a.
  2. Use χ 2 χ 2 test.
  3. Remember d.f. = K-2.

Table 3: TABLE 1
Actual Situation H o H o true H o H o false
decision accept Reject = error t. I reject Accept = error t. II
probability 1−α 1−α α α = significance level 1−β 1−β = power of the test β β

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