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Question
Statistics
Posted 5 months ago

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 nn chihuahuas and create a 99%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=zσnE = z \cdot \frac{\sigma}{\sqrt{n}}, where EE is the margin of error, zz is the z-score corresponding to the confidence level, σ\sigma is the standard deviation, and nn is the sample size
step 2
For a 99% confidence interval, the z-score (zz) is approximately 2.576
step 3
Given that the margin of error (EE) is 8 calories and the standard deviation (σ\sigma) is 30 calories, we substitute these values into the formula: 8=2.57630n8 = 2.576 \cdot \frac{30}{\sqrt{n}}
step 4
Solving for nn, we first isolate n\sqrt{n}: n=2.576308\sqrt{n} = \frac{2.576 \cdot 30}{8}
step 5
Calculate the value: n=77.288=9.66\sqrt{n} = \frac{77.28}{8} = 9.66
step 6
Square both sides to solve for nn: n=9.66293.31n = 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=94n = 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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