The amounts of vitamin C (in milligrams) for 100g (3.57 ounces) of various randomly selected fruits and vegetables are listed. Is there sufficient evidence to conclude that the standard deviation differs from 12 mg? Use alpha = 0.10. Use a graphing calculator

Answer:

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

Step-by-step explanation:

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, [tex]$ {\overset{-}X} =\frac{\Sigma (X)}{n} $[/tex]   =269.8/10=26.98

Sample Variance, [tex]$s^2= \frac{\Sigma(X-{\overset{-}X})^2}{(n-1)}$[/tex]    =1859.496/9=206.610667

Sample std dev,s=[tex]$\sqrt{s^2}$[/tex]=√206.610667≈14.373958

We want to test two tailed hypothesis:

[tex]$H_0$[/tex] : σ=12

[tex]$H_2$[/tex] : σ≠12

Given: s=14.373958⇒ [tex]$s^2$[/tex] =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 [tex]$ X_{1-\alpha/2}^2 =X_{0.95}^2 \text{ and}\ X_{1-\alpha/2}^2 =X_{0.95}^2 $[/tex]  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, [tex]$X_L^2$[/tex]=3.325  and [tex]$X_R^2$[/tex]=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 [tex]$H_0$[/tex]   if [tex]$x^2$[/tex]≤3.325  OR [tex]$x^2$[/tex]≥16.919

Test statistic is given by [tex]$X^2=\frac{(n-1)s^2}{\sigma^2}$[/tex] =((9)*206.610667)/144=12.913167    

Test statistic, [tex]$x^2$[/tex]=12.913

since test statistic is between two critical values, we (do not reject [tex]$H_0$[/tex]) (i.e, we fail to reject [tex]$H_0$[/tex].

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 [tex]$H_0$[/tex] 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 [tex]$H_0$[/tex]) (i.e, we fail to reject [tex]$H_0$[/tex]).

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

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