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;

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;

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;

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.