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, the researcher 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. the researcher would like to know if there is convincing evidence that the distribution of births is uniform across the days of the week. what is the value of the chi - square test statistic and the p - value of this test? find the chi square table here. ( chi^{2}=18.68 ), p - value is between 0.0025 and 0.005 ( chi^{2}=18.68 ), p - value is between 0.005 and 0.01 ( chi^{2}=24.70 ), p - value is less than 0.0005 ( chi^{2}=24.70 ), p - value is between 0.0005 and 0.001

Explanation:

Step1: Calculate the expected frequency

Since there are 7 days in a week and \(n = 350\) babies, the expected frequency \(E=\frac{350}{7}=50\) for each day.

Step2: Calculate the chi - square test statistic

The formula for the chi - square test statistic is \(\chi^{2}=\sum\frac{(O - E)^{2}}{E}\), where \(O\) is the observed frequency.
For Sunday: \(\frac{(28 - 50)^{2}}{50}=\frac{(- 22)^{2}}{50}=\frac{484}{50}=9.68\)
For Monday: \(\frac{(65 - 50)^{2}}{50}=\frac{15^{2}}{50}=\frac{225}{50}=4.5\)
For Tuesday: \(\frac{(48 - 50)^{2}}{50}=\frac{(-2)^{2}}{50}=\frac{4}{50}=0.08\)
For Wednesday: \(\frac{(56 - 50)^{2}}{50}=\frac{6^{2}}{50}=\frac{36}{50}=0.72\)
For Thursday: \(\frac{(45 - 50)^{2}}{50}=\frac{(-5)^{2}}{50}=\frac{25}{50}=0.5\)
For Friday: \(\frac{(62 - 50)^{2}}{50}=\frac{12^{2}}{50}=\frac{144}{50}=2.88\)
For Saturday: \(\frac{(46 - 50)^{2}}{50}=\frac{(-4)^{2}}{50}=\frac{16}{50}=0.32\)
\(\chi^{2}=9.68 + 4.5+0.08 + 0.72+0.5+2.88+0.32=18.68\)

Step3: Determine the degrees of freedom

The degrees of freedom \(df=k - 1\), where \(k = 7\) (number of days). So \(df=7-1 = 6\)

Step4: Find the P - value

Looking at the chi - square table with \(df = 6\), \(\chi^{2}=18.68\). The critical value \(\chi_{0.005,6}^{2}=18.548\) and \(\chi_{0.0025,6}^{2}=20.278\). So the \(P\) - value is between \(0.0025\) and \(0.005\)

Answer:

\(\chi^{2}=18.68\), \(P\) - value is between \(0.0025\) and \(0.005\)