A snack food manufacturer estimates that the variance of the

A snack food manufacturer estimates that the variance of the number of grams of carbohydrates in servings of its tortilla chips is1.21. A dietician is asked to test this claim and finds that a random sample of 19 servings has a variance of 1.05. At =0.10, is there enough evidence to reject the manufacturer\'s claim? Assume the population is normally distributed. Complete parts (a) through (e) below.

(a) Write the claim mathematically and identify H0 and Ha.
A.H0: 2 1.21
Ha:2 <1.21 (Claim)
B.H0: 2 1.21
Ha: 2 =1.21 (Claim)
C. H0: 2 =1.21 (Claim)
Ha: 2 1.21
D. H0: 2 1.21 (Claim)
   Ha: 2 >1.21

(b) Find the critical value(s) and identify the rejection region(s).
The critical value(s) is(are) ______.
(Round to three decimal places as needed. Use a comma to separate answers as needed.)

Choose the correct statement below and fill in the corresponding answer boxes.
A. The rejection region is 2 <_______.
B. The rejection region is 2 >______.
C. The rejection regions are 2<_____and 2greater than>_____.

(c) Find the standardized test statistic
2.________(Round to three decimal places as needed.)

(d) Decide whether to reject or fail to reject the null hypothesis, and (e) interpret the decision in the context of the original claim. Fill in the correct answers below.

a-Fail to reject
b- Reject

H0.

There
a- is not
b-is
enough evidence to at the 10% level of significance to reject the manufacturer\'s claim that the variance of the number of grams of carbohydrates in servings of its tortilla chips is 1.21.

Solution

A.H0: 2 1.21
Ha:2 <1.21 (Claim)

With = 10 and 19-1=18 df, the critical values are 25.989 and 28.869. Reject if < 25.989 or > 28.869.

The test statistic is:

=(19-1)*1.05/1.21=15.6

Reject the null hypothesis as 15.6<25.989.