STAT 201 Study Guide - Final Guide: Chi-Squared Test, Null Hypothesis, Dependent And Independent Variables
17 Jan 2019
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23. 14: ho: b1 = b2 = b3 = b4 = b5 = b6 = b7. If all the days are equally probable, then the probability would be 1/7 for each day and we would expect 100 births per day with 700 births in one week. X-squared = 18. 238, df = 6, p-value = 0. 005665. Chi-squared statistic: x^2 = sum of contribution from each cell (o e) ^2/ e. Chi-squared stat = 18. 238 df = k (possible outcomes) - 1 df = 6 (7-1) Using the chi-squared table, with df =6 and chi-squared = 18. 238, we get a p-value of 0. 001 < p < 0. 01 which is significant. Thus, 700 births are significant evidence that births are not equally probably on all. X-squared = 8. 8768, df = 6, p-value = 0. 1806. Df = 6 p-value = 0. 10 < p < 0. 20 which is non-significant. Thus, 700 births are not significant evidence that 20% of births occur on the weekends.
