Answer:
1. The chi-squared statistic = 10.36
The degrees of freedom = 17
The p-value for the test = 0.89
2. The range of the p-value from the Chi squared table = 0.75 < p-value < 0.90
Step-by-step explanation:
1. The Chi squared test is given as follows;
![\chi ^{2} = \sum \dfrac{\left (Observed - Expected \right )^{2}}{Expected }](https://tex.z-dn.net/?f=%5Cchi%20%5E%7B2%7D%20%3D%20%5Csum%20%5Cdfrac%7B%5Cleft%20%28Observed%20-%20Expected%20%20%5Cright%20%29%5E%7B2%7D%7D%7BExpected%20%20%7D)
Therefore,
UTI No UTI % Total
Cranberry juice 8 42 84 50
Lactobacillus 19 30 61 49
Control 18 30 60 50
The chi-squared statistic is given as follows;
![\chi ^{2} = \dfrac{\left (8- 18\right )^{2}}{18} + \dfrac{\left (42 - 30\right )^{2}}{30} = 10.36](https://tex.z-dn.net/?f=%5Cchi%20%5E%7B2%7D%20%3D%20%5Cdfrac%7B%5Cleft%20%288-%2018%5Cright%20%29%5E%7B2%7D%7D%7B18%7D%20%2B%20%20%5Cdfrac%7B%5Cleft%20%2842%20-%2030%5Cright%20%29%5E%7B2%7D%7D%7B30%7D%20%3D%2010.36)
The chi-squared statistic = 10.36
The degrees of freedom, df = 18 - 1 = 17 since the all of the expected count have a minimum value of 18
With the aid of the calculator we find the p value as p as follows;
![p = 0.9 - \dfrac{10.36 - 10.085}{12.972 - 10.085} \times (0.9 - 0.75)](https://tex.z-dn.net/?f=p%20%3D%200.9%20-%20%5Cdfrac%7B10.36%20-%2010.085%7D%7B12.972%20-%2010.085%7D%20%5Ctimes%20%280.9%20-%200.75%29)
The p-value for the test = 0.89
2. The range of the p-value from the Chi squared table is given as follows;
0.75 < p-value < 0.90.