Respuesta :
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;
[tex]\chi ^{2} = \sum \dfrac{\left (Observed - Expected \right )^{2}}{Expected }[/tex]
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;
[tex]\chi ^{2} = \dfrac{\left (8- 18\right )^{2}}{18} + \dfrac{\left (42 - 30\right )^{2}}{30} = 10.36[/tex]
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;
[tex]p = 0.9 - \dfrac{10.36 - 10.085}{12.972 - 10.085} \times (0.9 - 0.75)[/tex]
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.
