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