Chi-square Test and Exact Binomial Method for One Proportion

1. Functionalities

To determine if the population rate/proportion behind your data is significantly different from the specified rate/proportion

To determine how compatible the sample rate/proportion with a population rate/proportion

To determine the probability of success in a Bernoulli experiment

2. About your data

Your data come from binomial distribution (the proportion of success)

You know the whole sample and the number of specified events (the proportion of sub-group)

You have a specified proportion (p_{0})

Case Example

Suppose that in the general population, 20% of women who had infertility. Suppose a treatment may affect infertility. 200 women who were trying to get pregnant accepted the treatment.
Among 40 women who got the treatment, 10 were still infertile. We wanted to know if there was a significant difference in the rate of infertility among treated women compared to 20% of the general infertile rate.

Please follow the Steps, and Outputs will give real-time analytical results.

Output 1. Proportion Plot

Output 2. Test Results

1. Normal Theory Method with Yates' Continuity Correction, when np_{0}(1-p_{0}) >= 5

2. Exact Binomial Method, when np_{0}(1-p_{0}) < 5

Explanations

P Value < 0.05, then the population proportion/rate IS significantly different from the specified proportion/rate. (Accept the alternative hypothesis)

P Value >= 0.05, then the population proportion/rate IS NOT significantly different from the specified proportion/rate. (Accept the null hypothesis)

From the default settings, we concluded that there was no significant difference in the rate of infertility among homozygous women compared to the general infertility rate (P = 0.55). In this case, np_{0}(1-p_{0})=40*0.2*0.8 > 5, so the Normal Theory Method was preferable.

Chi-square Test for Two Independent Proportions

1. Functionalities

To determine if the population rate/proportion behind your 2 groups data are significantly different

2. About your data

Your 2 groups data come from binomial distribution (the proportion of success)

You know the whole sample and the number of specified events (the proportion of sub-group) from 2 groups

The 2 groups are independent observations

Case Example

Suppose all women in the study had at least on birth. We investigated 3220 breast cancer women as the case. Among them, 683 had at least one birth after 30 years old.
Also, we investigated 10245 no breast cancer women as control. Among them, 1498 had at least one birth after 30 years old.
We wanted to know if the underlying probability of having first birth over 30 years old was different in breast cancer and non-breast cancer groups.

Please follow the Steps, and Outputs will give real-time analytical results.

Output 1. Data Preview

Data Table

Percentage Plot of

1. Case

2. Control

Output 2. Test Results

Explanations

P Value < 0.05, then the population proportion/rate are significantly different in two groups. (Accept alternative hypothesis)

P Value >= 0.05, then the population proportion/rate are NOT significantly different in two groups. (Accept null hypothesis)

From the default settings, we conclude that women with breast cancer are significantly more likely to have their first child after 30 years old compared to women without breast cancer. (P<0.001)

Chi-square Test for More than Two Independent Proportions

1. Functionalities

To determine if the population rate/proportion behind your multiple group data are significantly different

2. About your data

Your group data come from binomial distribution (the proportion of success)

You know the whole sample and the number of specified events (the proportion of sub-group) from each group

The multiple groups are independent observations

Case Example

Suppose we wanted to study the relationship between age at first birth and the development of breast cancer. Thus, we investigated 3220 breast cancer cases and 10254 no breast cancer cases.
Then, we categorize women into different age groups.
We wanted to know if the probability of having cancer were different among different age groups; or if their ages related to breast cancer.

Please follow the Steps, and Outputs will give real-time analytical results.

P Value < 0.05, then the population proportion/rate are significantly different. (Accept the alternative hypothesis)

P Value >= 0.05, then the population proportion/rate are NOT significantly different. (Accept the null hypothesis)

In this default setting, we concluded that the probability of have cancer was significantly different in different age groups. (P < 0.001)

Chi-square Test for Trend in Multiple Independent Samples

1. Functionalities

To determine if the population rate/proportion behind your multiple group data vary

2. About your data

Your group data come from binomial distribution (the proportion of success)

You know the whole sample and the number of specified events (the proportion of sub-group) from each group

The multiple groups are independent observations

Case Example

Suppose we wanted to study the relationship between age at first birth and the development of breast cancer. Thus, we investigated 3220 breast cancer cases and 10254 no breast cancer cases.
Then, we categorize women into different age groups.
In this example, we wanted to know if the rate of having cancer tended from small to large ages.

Please follow the Steps, and Outputs will give real-time analytical results.