Randomly selected birth records were obtained, and categorized as listed in the table to the right. Use a 0.01 significance level to test the reasonable claim that births occur with equal frequency on the different days of the week. How might the apparent lower frequencies on Saturday and Sunday be explained?
Day
Number of Births
Sun
36
Mon
63
$\begin{array}{cc}\text { Tues } & \text { Wed } \\ 63 & 55\end{array}$
Wed 55
Thurs
61
Fri 55
Sat
37
Determine the null and alternative hypotheses.
$\mathrm{H}_{0}$ : Births occur with the same frequency on the different days of the week.
$H_{1}$ : At least one day has a different frequency of births than the other days.
Calculate the test statistic, $\chi^{2}$.
$\chi^{2}=15.454$ (Round to three decimal places as needed.)
Calculate the P-value.
P-value $=$
(Round to four decimal places as needed.)
Answer
Final Answer: The P-value is \(\boxed{0.017}\).
Step 1 ：Determine the null and alternative hypotheses.
Step 2 ：\(H_{0}\) : Births occur with the same frequency on the different days of the week.
Step 3 ：\(H_{1}\) : At least one day has a different frequency of births than the other days.
Step 4 ：Calculate the test statistic, \(\chi^{2}\).
Step 5 ：\(\chi^{2}=15.454\) (Round to three decimal places as needed.)
Step 6 ：Calculate the P-value.
Step 7 ：The P-value is the probability that a chi-square statistic having 6 degrees of freedom (7 days - 1) is more extreme than 15.454.
Step 8 ：The P-value is 0.017, which is greater than the significance level of 0.01. Therefore, we do not reject the null hypothesis. This means that we do not have enough evidence to support the claim that births occur with different frequencies on the different days of the week.
Step 9 ：Final Answer: The P-value is \(\boxed{0.017}\).
