A granola bar manufacturer states that their granola bars contain 28 grams of carbohydrates per bar.A sample of five granola bars is tested and a mean of 28.2g is obtained with a standard deviation of 2.7g.If carbohydrate content is normally distributed,the 95% confidence interval for the mean carbohydrate content of all such granola bars made by this company would be _____.

Question

Quizplus · Accepted Answer

We can use the formula for a confidence interval for the mean with a known population standard deviation:

CI = x̄ ± z(α/2) * (σ/√n)

where:
x̄ = sample mean = 28.2
σ = population standard deviation = 2.7
n = sample size = 5
z(α/2) = z-score for the desired confidence level

Since we want a 95% confidence interval, α = 0.05 and α/2 = 0.025. Looking up the z-score for 0.025 in a standard normal distribution table (or using a calculator), we find that z(0.025) = 1.96.

Substituting the values into the formula, we get:

CI = 28.2 ± 1.96 * (2.7/√5) ≈ [24.85, 31.55]

Therefore, the 95% confidence interval for the mean carbohydrate content of all granola bars made by this company is [24.85, 31.55]. The correct answer is (C).

A Granola Bar Manufacturer States That Their Granola Bars Contain