In: Statistics and Probability
Apply the χ2 goodness‐of‐fit test to the data in column 1 and test the assumption of a normal distribution.
Measured Force Data for Exercise Problems
F (N) Set 1 | F (N) Set 2 | F (N) Set 3 |
51.9 | 51.9 | 51.1 |
51.0 | 48.7 | 50.1 |
50.3 | 51.1 | 51.4 |
49.6 | 51.7 | 50.5 |
51.0 | 49.9 | 49.7 |
50.0 | 48.8 | 51.6 |
48.9 | 52.5 | 51.0 |
50.5 | 51.7 | 49.5 |
50.9 | 51.3 | 52.4 |
52.4 | 52.6 | 49.5 |
51.3 | 49.4 | 51.6 |
50.7 | 50.3 | 49.4 |
52.0 | 50.3 | 50.8 |
49.4 | 50.2 | 50.8 |
49.7 | 50.9 | 50.2 |
50.5 | 52.1 | 50.1 |
50.7 | 49.3 | 52.3 |
49.4 | 50.7 | 48.9 |
49.9 | 50.5 | 50.4 |
49.2 | 49.7 | 51.5 |
The Chi-Square goodness of fit test to test whether the normality assumption is true for column 1 is performed in following steps in excel.
Step 1: Write the data value in excel. The screenshot is shown below,
Step 2: The mean and standard deviation for the data values are obtained using the excel functions =AVERAGE() and STDEV(). The screenshot is shown below,
Step 3: The Z score for each data values are obtained using the formula,
The screenshot is shown below,
Step 4: The observed value are obtained by counting the Z score as follows,
The observed values are,
Observed values | |
Z score | Count |
Z<-1 | 4 |
-1<Z<0 | 5 |
0<Z<1 | 8 |
Z>1 | 3 |
Step 5: The expected values are obtained in excel. The screenshot is shown below,
The expected values are,
Expected values | |
Z score | Count |
Z<-1 | 3.173 |
-1<Z<0 | 6.827 |
0<Z<1 | 6.827 |
Z>1 | 3.173 |
Step 6: The Chi-Square statistic is obtained using the formula,
Z score | Observed, | Expected, | |||
Z<-1 | 4 | 3.173 | 0.827 | 0.684 | 0.215 |
-1<Z<0 | 5 | 6.827 | -1.827 | 3.338 | 0.489 |
0<Z<1 | 8 | 6.827 | 1.173 | 1.376 | 0.202 |
Z>1 | 3 | 3.173 | -0.173 | 0.030 | 0.009 |
Sum | 0.915 |
Step 7: The P-value for the chi square is obtained for chi square = 0.915 and degree of freedom = k - 1 = 4 - 1 = 3 using the excel function =1-CHISQ.DIST(0.915,3,TRUE)
Conclusion:
at 5% significance level. it can be concluded that the null hypothesis is not rejected. Hence we can conclude that the data values are normally distributed.