Question

Statistics

Posted 5 months ago

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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
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Answer from Sia

Posted 5 months ago

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.

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