In the following exercises we use the notation:

Problem1:According to one investigator's model the data are like 400 draws made at random from a large box. The null hypothesis says that the average of the box equals 50; the alternative says that the average of the box is more than 50. In fact, the data averaged out to 52.7 with and SD of 25. Compute z and P. What do you conclude? 
> observed_average := 52.7:
> expected_average := 50.:
> SE := 25./sqrt(400.):
> z := (observed_average  expected_average)/SE;
z := 2.160000000
The Pvalue is then the area under the normal curve from
2.16 to the right. This is the chance that the test statistic
will produce a value more extreme than the value observed: 2.16.
From the Normal table we have:
z Height Area z Height Area z Height Area ___________________ __________________ ___________________ 0.00 39.89 0.00 1.50 12.95 86.64 3.00 0.443 99.730 0.05 39.84 3.99 1.55 12.00 87.89 3.05 0.381 99.771 ..... 0.50 35.21 38.29 2.00 5.40 95.45 3.50 0.087 99.953 0.55 34.29 41.77 2.05 4.88 95.96 3.55 0.073 99.961 0.60 33.32 45.15 2.10 4.40 96.43 3.60 0.061 99.968 0.65 32.30 48.43 2.15 3.96 96.84 3.65 0.051 99.974 0.70 31.23 51.61 2.20 3.55 97.22 3.70 0.042 99.978 
> P := (100  96.84)/2;
P := 1.580000000
i.e. Pvalue is between 1% and 2%. This is a small chance so we reject the NULL in favor of the alternative that the average of the box is more than 50. 
Problem2:A coin is tossed 10,000 times, and it lands heads 5,167 times. Is the chance of heads equal to 50%? Or are there too many heads for that?

The tosses are like draws made at random from a box of tickets with zeroes and ones. 
> H0 := `Box contains an equal number of zeroes and ones`:
> H1 := `Box contains more ones than zeroes`:
another way of saying the same thing is; H0 = the proportion of ones in the box is 0.5. H1 = the proportion of ones in the box is more than 0.5. 
> observed_proportion := 5167./10000:
> expected_proportion := 5000./10000.:
> SE := sqrt(0.5*0.5)/sqrt(10000.):
> z := (observed_proportion  expected_proportion)/SE;
z := 3.340000000
z > 3 is a large value for standard units and we don't even
need to look at the table to conlude that P < half of 1%, since
we know that about 99% of the area under the normal curve is
between 3 and 3. If we want a more exact Pvalue we need
to look at the table again.
z Height Area z Height Area z Height Area ___________________ __________________ ___________________ .............. 0.25 38.67 19.74 1.75 8.63 91.99 3.25 0.203 99.885 0.30 38.14 23.58 1.80 7.90 92.81 3.30 0.172 99.903 0.35 37.52 27.37 1.85 7.21 93.57 3.35 0.146 99.919 0.40 36.83 31.08 1.90 6.56 94.26 3.40 0.123 99.933 0.45 36.05 34.73 1.95 5.96 94.88 3.45 0.104 99.944 
> P := (100  99.19)/2;
P := .4050000000
The exact Pvalue is then 0.405% or about 4 in 1000.
This is a very small chance and therefore we conclude that
there is strong evidence against the null. The coin is biased
towards heads.
The tTest
Problem3:Three measurements (in inches) of the length of a desk produce: 
If we assume the Gauss model for measurements we can think of the observations as 3 draws made at random from a box of tickets that follow the normal curve. The SD in the box is unknown and the number of draws is much less than 20 (i.e. only 3) so we need to use the ttest. First we compute the observed average and SD. 
> average := (50.8+50.9+50.7)/3.;
average := 50.80000000> SD := sqrt( ((50.8average)^2+(50.9average)^2+(50.7average)^2)/3.);
SD := .08164965809
what we need is to modify SD by multiplying it by the factor: sqrt(3/2), 
> SDplus := SD*sqrt(3./2.);
SDplus := .09999999996> t := (average  50)/(SDplus/sqrt(3.));
t := 13.85640647
This is an extremely large value for a zstatistic but not
so extreme for a tstatistic. We need to look at the ttable
with (31)=2 degrees of freedom.
Degrees (onetail areas) of freedom 10% 5% 1% ______________________________________ 1 3.08 6.31 31.82 2 1.89 2.92 6.96 3 1.64 2.35 4.54 4 1.53 2.13 3.75 5 1.48 2.02 3.36From the table we see that for 2 degrees of freedom the observed value t=13.8 is way to the right of 6.96 which is the cutoff point for 1%. Hence we conclude that Pvalue is much smaller that 1% and we reject the null hypothesis that the desk is 50 inches long. The desk is longer that 50 inches. 
Twosample ztest2 2 1/2 SEdifference := (SE1 + SE2 )Here is an example. Problem4:One hundred draws are made at random from box A, and 250 from box B. Both boxes contain a very large number of tickets. The sample from box A averages out to 220 with an SD of 70 and the sample from box B averages out to 180 with and SD of 50. Are the averages of the two boxes equal? What is the Pvalue? What is your conclusion? 
We just need to compute the zstatitic for the twosample test. For that we need the observed and expected differences and the SE for the difference. We have, 
> AVEa := 220 : SDa := 70 :
> AVEb := 180 : SDb := 50 :
> SEa := SDa/sqrt(100);
SEa := 7> SEb := SDb/sqrt(250);
1/2 SEb := 10> SEdifference := sqrt(SEa^2 + SEb^2);
1/2 SEdifference := 59> z := ( (AVEa  AVEb)  0 ) / SEdifference:
5.21
This is an extremely large value in standard units so we conclude
that the Pvalue is very close to 0 and we reject the null.
AVEa is in fact significantly bigger than AVEb.
Problem5:The National Household Survey on Drug Abuse was conducted in 1985 and in 1992.

For the marijuana case we have: 
> obsDiff := 21.9  11.0;
obsDiff := 10.9> SE85 := 100*sqrt(.219*(1..219))/sqrt(700.);
SE85 := 1.563142439> SE92 := 100*sqrt(.11*(1..11))/sqrt(700.);
SE92 := 1.182612121> SEdiff := sqrt(SE85^2+SE92^2);
SEdiff := 1.960098394> z := obsDiff/SEdiff;
z := 5.560945325
VERY LARGE z so VERY SMALL Pvalue so REJECT the null. Conclusion: Real drop in marijuana use from 1985 to 1992. For the cigarette case we have: 
> obsDiff := 36.031.9;
obsDiff := 4.1> SE85 := 100*sqrt(.36*(1..36))/sqrt(700.);
SE85 := 1.814229470> SE92 := 100*sqrt(.319*(1..319))/sqrt(700.);
SE92 := 1.761651011> SEdiff := sqrt(SE85^2+SE92^2);
SEdiff := 2.528802652> z := obsDiff/SEdiff;
z := 1.621320666
Now this is a very reasonable z value so there is no strong
evidence agains the null in this case. The pvalue is,
z Height Area z Height Area z Height Area ___________________ __________________ ___________________ 0.00 39.89 0.00 1.50 12.95 86.64 3.00 0.443 99.730 0.05 39.84 3.99 1.55 12.00 87.89 3.05 0.381 99.771 0.10 39.70 7.97 1.60 11.09 89.04 3.10 0.327 99.806 0.15 39.45 11.92 1.65 10.23 90.11 3.15 0.279 99.837 0.20 39.10 15.85 1.70 9.40 91.09 3.20 0.238 99.863 
> P := (10089.04)/2.;
P := 5.480000000
Pvalue larger than 5% so we can't reject the null hypothesis of
no difference in consumption of cigarettes in 1985 and 1992.
Randomized Controlled ExperimentsProblem6:Is Wheaties a power breakfast? To find out, a study is done in a large elementary statistics class; 499 students agree to participate; after the midterm, 250 are randomized to the treatment group, and 249 to the control group. The treatment group is fed Wheaties for breakfast 7 days a week. The control group gets Sugar Pops. The final scores averaged 66 for the treatment group; the SD was 21. For the control group, the average was 59 and the SD 20. What do you conclude? 
We proceed as if the treatment and control samples where independent samples with replacement from two separate boxes. Clearly that is not the case but the math turns out to be the same. This is due to the fact that the two mistakes approximately cancel each other out. Assuming that you are drawing WHITH replacement when in fact you are doing it WHITHOUT replacement inflates the SE but assuming that the two samples are INDEPENDENT when in fact they are DEPENDENT under estimates the SE the final result is that the SE turns out to be about right or a little inflated which makes the procedure a little overconservative but that's ok. For the above example we have, 
> H0 := `no difference between Wheaties and Sugar Pops`:
> H1 := `Wheaties are better`:
> observed_difference := 66  59:
> expected_difference := 0.0:
> SE_aveW := 21/sqrt(250.);
SE_aveW := 1.328156617> SE_aveSP := 20/sqrt(249.);
SE_aveSP := 1.267448501> SE_difference := sqrt(1.33^2 + 1.27^2);
SE_difference := 1.838967101> z := (observed_difference  expected_difference)/SE_difference;
z := 3.806484627
since z is almost 4 the Pvalue will be very small and will reject
the Null hypothesis that there is no difference between Wheaties and
Sugar Pops. Conclusion Wheaties are better and the Pvalue is:
z Height Area z Height Area z Height Area ___________________ __________________ ___________________ 0.00 39.89 0.00 1.50 12.95 86.64 3.00 0.443 99.730 .... 0.75 30.11 54.67 2.25 3.17 97.56 3.75 0.035 99.982 0.80 28.97 57.63 2.30 2.83 97.86 3.80 0.029 99.986 0.85 27.80 60.47 2.35 2.52 98.12 3.85 0.024 99.988 0.90 26.61 63.19 2.40 2.24 98.36 3.90 0.020 99.990 0.95 25.41 65.79 2.45 1.98 98.57 3.95 0.016 99.992 
> P := (100  99.986)/2.;
P := .007000000000
i.e. 0.007 percent! or 7 in 100,000 very very small indeed.
If the numbers where not hypothetical you should all be stuffing
yourselves of wheaties before the next exam!
The Chisquare Test 
> X2 := sum((obs_freq[i]  expt_freq[i])^2/expt_freq[i],i=1..K);
K  2 \ (obs_freq[i]  expt_freq[i]) X2 := )  / expt_freq[i]  i = 1
where K denotes the number of types of tickets in the box.
The Pvalue is obtained by looking at the area under the Chisquare
curve with (K1) degrees of freedom.
Here is an example:
Problem7:As part of a study on the selection of grand juries in Alameda county, the educational level of grand jurors was compared with the county distribution:Educational Number of level County Jurors _____________________________________________ Elemntary 28.4% 1 Secondary 48.5% 10 Some College 11.9% 16 College degree 11.2% 35 _______ ______ Total 100.0% 62Could a simple random sample of 62 people from the county show a distribution of educational level so different from the countywide one? Carry out the X2test and compute the Pvalue. 
> d1 := (1  62*.284)^2/(62*.284):
> d2 := (10  62*.485)^2/(62*.485):
> d3 := (16  62*.119)^2/(62*.119):
> d4 := (35  62*.112)^2/(62*.112):
> X2 := d1 + d2 + d3 + d4;
X2 := 152.4914064
This is way way way out in the tail of a X2 curve with 43 = 2 degrees of freedom. So reject with allllll confidence. 