Survey of 115 families chosen at random, having 5 children it was observed that the distribution was as follows
The 5 children are girls is 6
There is one boy and four girls is 18
There are two boys and three girls is 35
There are three boys and two girls is 30
There are four boys and one girl is 16
The five children are boys is 10
Question
Is there a significance difference between the observed frequency distribution and the binomial distribution with n =5 and p=0.05
Does the distribution follow a binomial propability distribution or there is something other than random variability causing the differences in the outcomes?
To determine whether there is a significant difference between the observed frequency distribution and the binomial distribution, we can perform a hypothesis test.
First, let's define the null and alternative hypotheses:
- Null hypothesis (H₀): The observed frequency distribution follows the binomial probability distribution with n = 5 and p = 0.05.
- Alternative hypothesis (H₁): There is something other than random variability causing the differences in the outcomes.
To test these hypotheses, we can use a chi-square goodness-of-fit test.
Step 1: Calculate the expected frequencies under the assumption that the null hypothesis is true. For a binomial distribution with n = 5 and p = 0.05, we can calculate the expected frequencies for each outcome.
Expected frequency for "The 5 children are girls" = (5 choose 5) * (0.05)^5 * (1-0.05)^(5-5) * 115
Expected frequency for "There is one boy and four girls" = (5 choose 4) * (0.05)^4 * (1-0.05)^(5-4) * 115
Expected frequency for "There are two boys and three girls" = (5 choose 3) * (0.05)^3 * (1-0.05)^(5-3) * 115
Expected frequency for "There are three boys and two girls" = (5 choose 2) * (0.05)^2 * (1-0.05)^(5-2) * 115
Expected frequency for "There are four boys and one girl" = (5 choose 1) * (0.05)^1 * (1-0.05)^(5-1) * 115
Expected frequency for "The five children are boys" = (5 choose 0) * (0.05)^0 * (1-0.05)^(5-0) * 115
Step 2: Conduct the chi-square test.
Calculate the chi-square statistic:
χ² = ∑ ((observed frequency - expected frequency)^2) / expected frequency
Step 3: Determine the critical value or p-value.
With (k-1) degrees of freedom, where k is the number of categories/outcomes (6 in this case), we can compare the calculated chi-square statistic to the critical value from the chi-square distribution table or calculate the p-value associated with the chi-square statistic using a chi-square distribution calculator.
Step 4: Make a decision.
If the calculated chi-square statistic is greater than the critical value or the p-value is lower than the significance level (usually 0.05), we reject the null hypothesis. This indicates a significant difference between the observed and expected frequencies, implying that the distribution does not follow a binomial probability distribution.
If the calculated chi-square statistic is smaller than the critical value or the p-value is higher than the significance level, we fail to reject the null hypothesis. This suggests that there is not enough evidence to conclude that the distribution significantly deviates from the binomial distribution assumption.
By following these steps, you can perform the chi-square goodness-of-fit test to determine whether there is a significant difference between the observed frequency distribution and the binomial distribution.