Questions: A snack food manufacturer estimates that the variance of the number of grams of carbohydrates in servings of its tortilla chips is 1.25. A dietician is asked to test this claim and finds that a random sample of 24 servings has a variance of 1.27. At α=0.05, 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: σ^2 ≥ 1.25
B. H0: σ^2 ≠ 1.25 Ha: σ^2<1.25 (Claim) Ha: σ^2=1.25 (Claim)
C. H0: σ^2 ≤ 1.25 (Claim) D. Ha: σ^2>1.25
H0: σ^2=1.25 (Claim)
Ha: σ^2 ≠ 1.25
(b) Find the critical value(s) and identify the rejection region(s).
Transcript text: A snack food manufacturer estimates that the variance of the number of grams of carbohydrates in servings of its tortilla chips is 1.25. A dietician is asked to test this claim and finds that a random sample of 24 servings has a variance of 1.27. At $\alpha=0.05$, 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 $\mathrm{H}_{0}$ and $\mathrm{H}_{a}$.
A. $H_{0}: \sigma^{2} \geq 1.25$
B. $H_{0}: \sigma^{2} \neq 1.25$ $H_{a}: \sigma^{2}<1.25$ (Claim) $\mathrm{H}_{\mathrm{a}}: \sigma^{2}=1.25$ (Claim)
C. $H_{0}: \sigma^{2} \leq 1.25$ (Claim) D. $H_{a}: \sigma^{2}>1.25$
\[
\begin{array}{l}
H_{0}: \sigma^{2}=1.25 \text { (Claim) } \\
H_{a}: \sigma^{2} \neq 1.25
\end{array}
\]
(b) Find the critical value(s) and identify the rejection region(s).
Solution
Solution Steps
Step 1: Calculate the Test Statistic
To test the manufacturer's claim regarding the variance of the number of grams of carbohydrates in servings of tortilla chips, we calculate the test statistic using the formula for the chi-square statistic:
Next, we find the P-value corresponding to the calculated test statistic. The P-value is determined from the chi-square distribution with \( n - 1 = 23 \) degrees of freedom:
\[
P = P(\chi^2(23) \geq 23.368) = 0.8789
\]
Step 3: Identify Critical Values and Rejection Regions
For a two-tailed test at a significance level of \( \alpha = 0.05 \), we find the critical values from the chi-square distribution:
Since the calculated test statistic \( \chi^2 = 23.368 \) does not fall into the rejection regions (i.e., \( 11.6886 < 23.368 < 38.0756 \)), we fail to reject the null hypothesis \( H_0: \sigma^2 = 1.25 \).
Final Answer
The conclusion is that there is not enough evidence to reject the manufacturer's claim regarding the variance. Thus, the answer is: