In: Statistics and Probability
A doctor claims that the number of births by day of the week is
uniformly distributed. To test the claim, you randomly select
births from a recent year and record the day of the week the birth
takes place. Use a 1% significance to test the claim.
Round to the fourth as needed.
Day | Sun | Mon | Tues | Wed | Thurs | Fri | Sat |
Frequency | 56 | 36 | 31 | 42 | 36 | 48 | 52 |
a) Type out the null and alternative hypothesis
b) State you test statistic
c) State your p-value
d) State your decision rule
e) Type out your conclusion
a)
null hypothesis: Ho: number of births by day of the week is uniformly distributed
Alternate hypothesis: Ha: number of births by day of the week is not uniformly distributed
b)
Applying chi square test:
relative | observed | Expected | residual | Chi square | |
category | frequency | Oi | Ei=total*p | R2i=(Oi-Ei)/√Ei | R2i=(Oi-Ei)2/Ei |
Sun | 1/7 | 56.000 | 43.00 | 1.98 | 3.930 |
Mon | 1/7 | 36.000 | 43.00 | -1.07 | 1.140 |
Tue | 1/7 | 31.000 | 43.00 | -1.83 | 3.349 |
Wed | 1/7 | 42.000 | 43.00 | -0.15 | 0.023 |
Thu | 1/7 | 36.000 | 43.00 | -1.07 | 1.140 |
Fri | 1/7 | 48.000 | 43.00 | 0.76 | 0.581 |
Sat | 1/7 | 52.000 | 43.00 | 1.37 | 1.884 |
total | 1.000 | 301 | 301 | 12.047 |
test statisitc X2 =12.047
c)
p value =0.0609
d)
Decision rule: reject HO if p value <0.01 or tst statistic X2 <16.812
e)
as test statistic is not in rejection region ; we can not conclude that number of births by day of the week is not uniformly distributed