Are births evenly distributed across days of the week? A random sample of births

Question

Are births evenly distributed across days of the week? A random sample of births from a local health department gave the folowing information Day Dirths (a) If births are evenly distributed, how many would you expect on each day? (one decimal place) (b) Would a goodness of fit test be valid in this situation? o Yes. The expected count for each day is at least S, and the sample is random Yes. The sample sizc is largc cnough, and the samplc is random. No. There are two days with less than 5 births. (c) Open datafile Births and Days of Week in StatCrunch to find the test statistic and P-value. At the 0.01 significance level, what would you conclude? (If you need help with StatCrunch, watch this video https://www.screencast.com/t/QiS3cdwQN2) cast.com/t/Qi53 Do not reject the null hypothesis. There is sufficient ovidence to indicate that births are unevenly distributed over the days of the weck o Do not reject the null hypothesis. There is not enough evidence to indicate that births are not evenly distributed over the days of the week. o Reject the null hypothesis. There is insufficient evidence to indicate that births are unevenly distributed over the days of the week. e Reject the nul hypothesis. There is evidence to indicate that births are not evenly distributed over the days of the week

Explanation / Answer

a) there are all total 54 births in 7 days.

so if the births were uniformly distributed then expected number of births each day is 54/7=7.7  

b) yes. the sample is random. and from part a) the expected count for each day is 7.7 which is greater than 5.

hence chi square test for goodness of fit is valid

c) there are 7 days. let fi denotes the number of births for ith day i=1,2,3,4,5,6,7

so f1=3 f2=7 f3=12 f4=10 f5=9 f6=11 f7=2

here sample size=n=54

if the data were from uniform distribution then the proportion pi for ith day wil be pi=1/7 i=1,2,3,4,5,6,7

so H0: p1=p2=p3=p4=p5=p6=p7=1/7 H1:not H0

the test statistic is T=(f1-n*p1)2/(n*p1)+.....+(f7-n*p7)2/(n*p7) which under H0 follows a chi square distribution with df=7-1=6

so the test statistic is T=(3-7.7)2/7.7+(7-7.7)2/7.7+(12-7.7)2/7.7+(10-7.7)2/7.7+(9-7.7)2/7.7+(11-7.7)2/7.7+(2-7.7)2/7.7 =11.874

so p value is P[T>11.874] where T follows chi square with df 6

so p=0.06483947 [answer]

level of significace=0.01

so p>0.01

hence the conclusion is

do not reject null hypothesis. there is not enough evidence to indicate that births are not evenly distributed over the days of the week

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