Question

QUESTION 1

1. A random group of oranges were selected from an orchard to analyze their ripeness. The data is shown

   below:

   Ready to pick

   Ripe

   Ready in three days

   Ready in one week

   Ready in two weeks

   Number of oranges

   11

   20

   13

   17

   Based on the time of year, the orchard owner believes that 30% of the oranges are ready for picking now,

   30% will be ready in three days, 30% will be ready in one week, and 10% will be ready in two weeks.
   Is there evidence to reject this hypothesis at a  .05 significance level?  

   		There is evidence to reject the claim that the oranges are distributed as claimed because the test value 24.082 > 9.488
   		There is not evidence to reject the claim that the oranges are distributed as claimed because the test value 5.991 < 24.082
   		There is evidence to reject the claim that the oranges are distributed as claimed because the test value 24.082 > 7.815
   		There is not evidence to reject the claim that the oranges are distributed as claimed because the test value 9.488 < 24.082

Answer 1

Chi-square test of Goodness of fit is a test of significance to test whether the experimental result support the theoretical results or not.

Under H0, we check if there is any difference between observed and expected frequencies sum(Oi)=N=sum(Ei)
p=number of parameters estimated for fitting the distribution

The chi-square test statistic is given as:

Conditions for applying goodness of fit test:

• sample observations should be independent
• total frequency should be greater than 50.
• each of the expected frequency should be greater than or equal to 5.

Decision Rule:

R-commands and outputs:

> Oi=c(11,20,13,17) ## Observed frequency
> N=sum(Oi) ## total frequency is greater than 50
> N
[1] 61
> pi=c(30/100,30/100,30/100,10/100)
> Ei=N*pi ## Expected frequency # Each Ei is greater than 5.
> Ei
[1] 18.3 18.3 18.3 6.1
> (Oi-Ei)^2/Ei
[1] 2.9120219 0.1579235 1.5349727 19.4770492
> sum((Oi-Ei)^2/Ei) ## Calculated value of test statistic
[1] 24.08197

> qchisq(1-0.05,df=4-1) ## Tabulated (or Critical) value at alpha percent level of significance ## k=3=number of classes ## Here, p=0
[1] 7.814728
##7.814728 7.815

### There is evidence to reject the claim that the oranges are distributed as claimed because the test value 24.082 > 7.815.