Dr. Suarez, a veterinarian, wants to estimate the average number of calories a chihuahua typically eats in a day. She'll take a sample of $n$ chihuahuas and create a $99 \%$ confidence interval for the mean daily caloric intake. She wants the margin of error to be no more than 8 calories. A pilot study suggests that the daily caloric intake of chihuahuas has a standard deviation of 30 calories.

Which of these is the smallest approximate sample size required to obtain the desired margin of error?

Choose 1 answer:
(A) 8 chihuahuas
(B) 30 chihuahuas
(c) 55 chihuahuas

D 94 chihuahuas
(E) 115 chihuahuas

Question

Dr. Suarez, a veterinarian, wants to estimate the average number of calories a chihuahua typically eats in a day. She'll take a sample of $n$ chihuahuas and create a $99 \%$ confidence interval for the mean daily caloric intake. She wants the margin of error to be no more than 8 calories. A pilot study suggests that the daily caloric intake of chihuahuas has a standard deviation of 30 calories.

Which of these is the smallest approximate sample size required to obtain the desired margin of error?

Choose 1 answer:
(A) 8 chihuahuas
(B) 30 chihuahuas
(c) 55 chihuahuas

D 94 chihuahuas
(E) 115 chihuahuas

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To determine the smallest sample size required for the desired margin of error, we use the formula for the margin of error in a confidence interval for the mean: $E = z \cdot \frac{\sigma}{\sqrt{n}}$, where $E$ is the margin of error, $z$ is the z-score corresponding to the confidence level, $\sigma$ is the standard deviation, and $n$ is the sample size. ‖ ‖ step 2 ⋮ For a 99% confidence interval, the z-score ($z$) is approximately 2.576. ‖ ‖ step 3 ⋮ Given that the margin of error ($E$) is 8 calories and the standard deviation ($\sigma$) is 30 calories, we substitute these values into the formula: $8 = 2.576 \cdot \frac{30}{\sqrt{n}}$. ‖ ‖ step 4 ⋮ Solving for $n$, we first isolate $\sqrt{n}$: $\sqrt{n} = \frac{2.576 \cdot 30}{8}$. ‖ ‖ step 5 ⋮ Calculate the value: $\sqrt{n} = \frac{77.28}{8} = 9.66$. ‖ ‖ step 6 ⋮ Square both sides to solve for $n$: $n = 9.66^2 \approx 93.31$. ‖ ‖ step 7 ⋮ Since the sample size must be a whole number, we round up to the next whole number: $n = 94$. ‖ ∻Answer∻ ***D*** ∻Key Concept∻ ⚹Sample size calculation for a confidence interval⚹ ∻Explanation∻ ⚹To achieve a specific margin of error in a confidence interval, the sample size must be calculated using the formula for the margin of error, considering the z-score for the desired confidence level and the standard deviation from a pilot study.⚹◊What is the formula for calculating the sample size needed to obtain a desired margin of error in a confidence interval for the mean⍭ Generate me a similar question◊

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