The following 40 data points were collected representing rea

The following 40 data points were collected representing reaction times (in sec) during a psychological experiment:

     2.68      3.41     3.44     3.45     3.74     3.84     3.87     4.10

     4.15      4.21     4.25     4.29     4.40     4.40     4.45     4.63

     4.64      4.65     4.74     4.84     4.89     4.90     4.94     4.99

     5.05      5.06     5.21     5.22     5.26     5.28     5.32     5.52

     5.55      5.59     5.80     6.10     6.16     6.51     6.93     7.08

1) Using the chi-square goodness-of-fit test, test the hypothesis that the data is normally distributed using a = 0.05.

2) Using the K-S goodness-of-fit test, test the hypothesis that the data is normally distributed using a = 0.05.

Solution

Null Hypothesis H0:Sample is normal distributed                          

vs H1:Sample is not normally distributed                            

Bin         Frequency(Observed)   Frequency(Expected)    (observed-expected)^2/(expected)

0-3          1              1.194717827       0.031735554

3-3.5      3              2.214618475       0.278523884

3.5-4      3              4.401062383       0.446023171

4-4.5      8              6.76676263          0.224756578

4.5-5      9              8.050013523       0.112108421

5-5.5      7              7.409957965       0.022681037

5.5- 6     4              5.277557622       0.309263033

6- 6.5     2              2.908215328       0.283629301

6.5- 7     2              1.239820979       0.466093213

7-7.5      1              0.537273268       0.39852351

                40           40           2.573337703

Degre of freedom is 7                   

*             3 degree of freedom is lost as two parameter is estimated        

1)            Mean    4.8385

2)            Sample Standard deviation         0.953564299

At 5% level of signifance we have insuffient evidence to reject null hypothesis since critical value is very less as compared to critical value chi-square distribution with DOF 7 i.e-14.07               

(b)

Null Hypothesis H0:Sample is normal distributed                                                                          

vs H1:Sample is not normally distributed                                                                            

Bin         Frequency          Frequency cumulative Sn           Z-score F(x)        Difference

3              1              1              0.025     -1.928029397      0.026926              0.001926

3.5          3              4              0.1          -1.403680907      0.080207              0.019793

4              3              7              0.175     -0.879332417      0.189611              0.014611

4.5          8              15           0.375     -0.354983928      0.361301              0.013699

5              9              24           0.6          0.169364562       0.567245              0.032755

5.5          7              31           0.775     0.693713052       0.756069              0.018931

6              4              35           0.875     1.218061542       0.8884   0.0134

6.5          2              37           0.925     1.742410032       0.959282              0.034282

7              2              39           0.975     2.266758522       0.988298              0.013298

7.5          1              40           1              2.791107011       1              0

                                Dn=max               0.034282                                                                                                                             

                                Dn,alpha              0.215035                                                                                                                             

Since Dn = 0.034282 < 0.215035 = Dn,, we conclude that the data is a good fit with the normal distribution.