Assume that the readings at freezing on a batch of thermometers are normally distributed with a mean of 0°C
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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.