Question

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 1

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

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,
`$ {\overset{-}X} =(\Sigma (X))/(n) $` =269.8/10=26.98

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

Sample std dev,s=
`$√(s^2)$`=√206.610667≈14.373958

We want to test two tailed hypothesis:

`$H_0$` : σ=12

`$H_2$` : σ≠12

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

Test statistic is given by
`$X^2=((n-1)s^2)/(\sigma^2)$` =((9)*206.610667)/144=12.913167

Test statistic,
`$x^2$`=12.913

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

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
`$H_0$` 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
`$H_0$`) (i.e, we fail to reject
`$H_0$`).

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