determined. Twenty six bottles of Brand A yielded an average fat content of 26 grams with a standard
deviation of 5 grams, and 31 bottles of Brand B yielded an average fat content of 25.8 grams with a
standard deviation of 7 grams. Test the hypothesis at the level of significance that the variance
are different. (15)

Question

determined. Twenty six bottles of Brand A yielded an average fat content of 26 grams with a standard
deviation of 5 grams, and 31 bottles of Brand B yielded an average fat content of 25.8 grams with a
standard deviation of 7 grams. Test the hypothesis at the

level of significance that the variance
are different. (15)

UpStudy AI · Accepted Answer

We wish to test whether the fat‐content variances for Brand A and Brand B are equal. The information provided is: • Brand A: sample size n₁ = 26, sample mean = 26 grams, standard deviation s₁ = 5 grams (so variance s₁² = 25). • Brand B: sample size n₂ = 31, sample mean = 25.8 grams, standard deviation s₂ = 7 grams (so variance s₂² = 49). Because we are testing whether the variances are different (a two‐tailed test) we set up the hypotheses as follows: H₀: σ_A² = σ_B² (the variances are equal) Hₐ: σ_A² ≠ σ_B² (the variances differ) Step 1. Choice of Test Statistic We use the F-test for equality of variances. To keep the statistic F ≥ 1, we define F = (larger sample variance) / (smaller sample variance). Here, s₂² (49) is larger than s₁² (25). Thus: F = 49 / 25 = 1.96. The degrees of freedom are numerator d.f. = n₂ – 1 = 31 – 1 = 30, denominator d.f. = n₁ – 1 = 26 – 1 = 25. Step 2. Significance Level and Rejection Regions We are testing at a 10% significance level (α = 0.10). Since the test is two‐tailed, the significance in each tail is 0.05. The rejection region for the F-test is: Reject H₀ if F > F₍₀.₉₅₎(30,25) or F < F₍₀.₀₅₎(30,25). It is often easier to use the reciprocal relationship: F₍₀.₀₅₎(30,25) = 1 / F₍₀.₉₅₎(25,30). Using standard F–tables (or an appropriate calculator) one obtains approximate critical values. For example, one set of approximations might be: F₍₀.₉₅₎(30,25) ≈ 1.84 and F₍₀.₀₅₎(30,25) ≈ 1/ F₍₀.₉₅₎(25,30) ≈ 0.56. Thus, the rejection region is: Reject H₀ if F > 1.84 or F < 0.56. Step 3. Decision Our computed F-statistic is 1.96. Since 1.96 > 1.84, it falls in the rejection region. Step 4. Conclusion At the 10% level of significance, there is sufficient evidence to reject the null hypothesis that the variances are equal. We therefore conclude that the fat content variances of Brand A and Brand B are significantly different. Note: The exact critical values depend on the F–distribution tables or software used. However, with our approximations the conclusion remains the same. At the 10% significance level, the variances of Brand A and Brand B are significantly different.

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