Question

a researcher would like to investigate if the distribution of the number of babies born is uniform across the days of the week. to investigate, he selects a random sample of 350 babies and records on which day of the week they were born. he finds that 28 babies were born on sundays, 65 on mondays, 48 on tuesdays, 56 on wednesdays, 45 on thursdays, 62 on fridays, and 46 on saturdays. he tests the following hypotheses: ( h_0 ): the distribution of births is uniform across the days of the week. ( h_a ): the distribution of births is not uniform across the days of the week. the chi - square test statistic is ( chi^2 = 18.68 ) and the p - value is between 0.0025 and 0.005. because the p - value is less than 0.05, the researcher rejects the null hypothesis. because the null hypothesis is rejected, the researcher would like to know which term contributed the significance of the test. complete a follow - up analysis. which term contributed the most to the significance of this test? were there more or fewer births on this day than expected? sunday, more births than expected sunday, fewer births than expected monday, more births than expected monday, fewer births than expected

Explanation:

Step1: Calculate the expected number of births per day

Since there are 7 days in a week and \(n = 350\) babies, the expected number of births per day \(E=\frac{350}{7}=50\)

Step2: Calculate the \(\chi^{2}\) contribution for each day

The formula for the \(\chi^{2}\) contribution of a category is \(\frac{(O - E)^{2}}{E}\), where \(O\) is the observed value and \(E\) is the expected value.

• For Sunday: \(O = 28\), \(E=50\). The contribution is \(\frac{(28 - 50)^{2}}{50}=\frac{(- 22)^{2}}{50}=\frac{484}{50}=9.68\)
• For Monday: \(O = 65\), \(E = 50\). The contribution is \(\frac{(65 - 50)^{2}}{50}=\frac{15^{2}}{50}=\frac{225}{50}=4.5\)
• For Tuesday: \(O = 48\), \(E = 50\). The contribution is \(\frac{(48 - 50)^{2}}{50}=\frac{(-2)^{2}}{50}=\frac{4}{50}=0.08\)
• For Wednesday: \(O = 56\), \(E = 50\). The contribution is \(\frac{(56 - 50)^{2}}{50}=\frac{6^{2}}{50}=\frac{36}{50}=0.72\)
• For Thursday: \(O = 45\), \(E = 50\). The contribution is \(\frac{(45 - 50)^{2}}{50}=\frac{(-5)^{2}}{50}=\frac{25}{50}=0.5\)
• For Friday: \(O = 62\), \(E = 50\). The contribution is \(\frac{(62 - 50)^{2}}{50}=\frac{12^{2}}{50}=\frac{144}{50}=2.88\)
• For Saturday: \(O = 46\), \(E = 50\). The contribution is \(\frac{(46 - 50)^{2}}{50}=\frac{(-4)^{2}}{50}=\frac{16}{50}=0.32\)

Answer:

Sunday, fewer births than expected