A nutritionist wants to study the effects of storage time (6, 12, & 18 months) on the amount of vitamin C present
in freeze dried fruit. Vitamin C is measured in milligrams
per 100 milligrams of fruit. Six fruit packs were randomly
assigned to each of the three storage times.
The treatment is ?

The values are missing in the question. They are : 16.3,12.8,13,32.2,28.1,34.4,46.4,53,15.4,18.2

16.3,12.8,13,32.2,28.1,34.4,46.4,53,15.4,18.2

We calculate sample mean and std deviation from given data.

Sample Mean,    =269.8/10=26.98

Sample Variance,     =1859.496/9=206.610667

Sample std dev,s==√206.610667≈14.373958

We want to test two tailed hypothesis:

: σ=12

: σ≠12

Given: s=14.373958⇒  =206.610667,n=10

Significance level, a=0.1 Degrees of freedom, df=n-1=9

Since this is two tailed test, we need area of α=0.1 on either side of critical value. This means area of 0.05 on right of positive critical value.

Critical values are given by   with 9 degrees of freedom.

We can use chi-square table or following excel commands:

CHISQ. INV(0.95,9)=3.32511284307

CHISQ. INV (0.05,9)=16.9189776046

Critical Values, =3.325  and =16.919

Since this is two tailed test, the decision rule or rejection criteria is:

Reject null hypothesis if test statistic is less than or equal to left critical value OR more than or equal to right critical value, i.e,

Reject    if ≤3.325  OR ≥16.919

Test statistic is given by  =((9)*206.610667)/144=12.913167    

Test statistic, =12.913

since test statistic is between two critical values, we (do not reject ) (i.e, we fail to reject .

Conclusion: There is (not sufficient) evidence to conclude that standard deviation is not equal to 12 .

Alternatively, we can use p-value approach. The decision rule or rejection criteria is: reject null hypothesis if p-value is less than or equal to significance level (α), i.e, reject  if p-value ≤0.1

P-value is twice the (smaller) tail area of test statistic because this is two tailed test. We can use excel function to find this. 2*M IN (CHISQ.DIST (12.913167,9,  TRUE ) 1-CHISQ.DIST 12.913167,9,  TRUE ) =0.333149809868

P-value =0.3331

since p-value is greater than significance level (α),we (do not reject ) (i.e, we fail to reject ).

Conclusion: There is (not sufficient) evidence to conclude that standard deviation is different from 12 .

the f

last graph i believe

step-by-step explanation:

worked it out i believe