Calculate the test statistic (Round to three decimal places as needed.)Calculate the P-value. (Round to four decimal places as needed.)conclusion 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 ofthe week. How might the apparent lower frequencies on Saturday and Sunday be explained?DaySunMonTuesWedThursFriSatNumber of Births50636362576446Determine the null and alternative hypotheses.Ho: Births occur with the same frequency on the different days of the week.H1: At least one day has a different frequency of births than the other days.

Test statistic: The number of days is 7. Let us consider, the proportion as 1/7 for each day. Consider, the expected frequencies, Number of Births(O) Proportion Expected frequency=Total of observed frequency*proportion (O-E)^2/E 50 0.142857143 57.85714286 1.067019 63 0.142857143 57.85714286 0.457143 63 0.142857143 57.85714286 0.457143 62 0.142857143 57.85714286 0.296649 57 0.142857143 57.85714286 0.012698 64 0.142857143 57.85714286 0.652205 46 0.142857143 57.85714286 2.429982 Total=405 ∑(O-E)2E=5.37284 Test statistic, χ2=∑(O-E)2E =5.373, rounded to three decimalsP value: df=number of days-1=7-1=6 P value=0.4969, from the excel function, =CHISQ.DIST.RT(5.373,6) Decision Rule: If p-value ≤ α, then reject the null hypothesis. Conclusion: The level of significance is α=0.01 Here, the p-value is greater than the level of significance. From the decision rule, failed to reject the null hypothesis. Conclusion is “Births occur with the same frequency on the different days of the week”.