Question

The following data were collected related to the number of visits to a suburban health clinic.

Month	Vists
January	1610
February	1585
March	1649
April	1590
May	1540
June	1397
July	1410
August	1350
September	1495
Ocotober	1564
November	1602
December	1655

The null hypothesis is that the number of visits is uniformly distributed throughout the year. Use a chi-square goodness-of-fit test to determine whether the null hypothesis should be rejected at α = 0.05. Interpret the results.

Answer 1

null hypothesis:Ho:number of visits are uniformly distributed.through out the year

Alternate hypothesis:Ho:number of visits are not uniformly distributed.through out the year

degree of freedom =categories-1=			11
for 0.05 level and 11 df :crtiical value X2 =				19.675
Decision rule: reject Ho if value of test statistic X2>19.675
applying chi square goodness of fit test:

	relative	observed	Expected	residual	Chi square
category	frequency(p)	Oi	Ei=total*p	R2i=(Oi-Ei)/√Ei	R2i=(Oi-Ei)2/Ei
Jan	1/12	1610.000	1537.25	1.86	3.443
Feb	1/12	1585.000	1537.25	1.22	1.483
Mar	1/12	1649.000	1537.25	2.85	8.124
Apr	1/12	1590.000	1537.25	1.35	1.810
May	1/12	1540.000	1537.25	0.07	0.005
Jun	1/12	1397.000	1537.25	-3.58	12.796
Jul	1/12	1410.000	1537.25	-3.25	10.533
Aug	1/12	1350.000	1537.25	-4.78	22.809
Sep	1/12	1495.000	1537.25	-1.08	1.161
Oct	1/12	1564.000	1537.25	0.68	0.465
Nov	1/12	1602.000	1537.25	1.65	2.727
Dec	1/12	1655.000	1537.25	3.00	9.019
total	1.000	18447	18447		74.3758
test statistic X2 =		74.376

since test statistic falls in rejection region we reject null hypothesis

we have sufficient evidence to conclude that number of visits are not uniformly distributed.through out the year