Question

a snack food manufacturer estimates that the variance of the number of grams of carbohydrates in servings of its tortilla chips is 1.14. a dietician is asked to test this claim and finds that a random sample of 24 servings has a variance of 1.14. at α = 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: σ² ≠ 1.14
hₐ: σ² = 1.14 (claim)

b. h₀: σ² ≥ 1.14
hₐ: σ² < 1.14 (claim)

c. h₀: σ² ≤ 1.14 (claim)
hₐ: σ² > 1.14

d. h₀: σ² = 1.14 (claim)
hₐ: σ² ≠ 1.14

(b) find the critical value(s) and identify the rejection region(s).

the critical value(s) is(are) .
(round to two decimal places as needed. use a comma to separate answers as needed.)

Explanation:

Step1: Identify the distribution and parameters

We are testing a hypothesis about the variance of a normally distributed population, so we use the chi - square distribution. The sample size \(n = 24\), so the degrees of freedom \(df=n - 1=24 - 1 = 23\). The significance level \(\alpha=0.10\). Since the alternative hypothesis \(H_{a}:\sigma^{2}
eq1.14\) (from part (a), option D), this is a two - tailed test. So we split the \(\alpha\) into two tails, \(\alpha/2=0.05\) and \(1-\alpha/2 = 0.95\).

Step2: Find the critical values

We need to find \(\chi_{1 - \alpha/2}^{2}\) and \(\chi_{\alpha/2}^{2}\) with \(df = 23\).

• For \(\chi_{0.95}^{2}\) (the lower critical value) with \(df = 23\), using the chi - square distribution table or a calculator function (e.g., in a TI - 84 Plus, invChi2(0.95,23)), we get \(\chi_{0.95}^{2}\approx13.09\).
• For \(\chi_{0.05}^{2}\) (the upper critical value) with \(df = 23\), using the chi - square distribution table or a calculator function (e.g., invChi2(0.05,23)), we get \(\chi_{0.05}^{2}\approx35.17\).

Answer:

13.09, 35.17