A nutritionist wants to estimate the average daily calcium intake (in mg\text{mg}) among teenagers in a city. Previous studies report the population standard deviation as σ=120 mg\sigma=120\text{ mg}. What is the minimum sample size required to estimate the true mean within ±30 mg\pm30\text{ mg} with 99%99\% confidence? Additionally, explain briefly why increasing the confidence level from 95%95\% to 99%99\% increases the required sample size, assuming the same margin of error and population standard deviation.

n=107n=107Because of the higher zz-value for 99%99\% confidence level

n=112n=112Because of the lower zz-value for 95%95\% confidence level

n=98n=98Because of the higher standard error at 95%95\% confidence level

n=107n=107Because of the lower standard error at 95%95\% confidence level

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1. Intro to Stats and Collecting Data(0)

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Visualizing Qualitative vs. Quantitative Data
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Frequency Distributions
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Interpreting Standard Deviation
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Sampling Distribution of the Sample Mean and Central Limit Theorem
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Introduction to Confidence Intervals
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Confidence Intervals for Population Mean
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Determining the Minimum Sample Size Required
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7. Sampling Distributions & Confidence Intervals: Mean

Determining the Minimum Sample Size Required

7. Sampling Distributions & Confidence Intervals: Mean

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Problem 5Multiple Choice

A nutritionist wants to estimate the average daily calcium intake (in $mg$ ) among teenagers in a city. Previous studies report the population standard deviation as $sigma equals 120 mg$ . What is the minimum sample size required to estimate the true mean within $\pm 30 mg$ with $99 percent sign$ confidence? Additionally, explain briefly why increasing the confidence level from $95 percent sign$ to $99 percent sign$ increases the required sample size, assuming the same margin of error and population standard deviation.

$n equals 112$

Because of the lower $z$ -value for $95 percent sign$ confidence level

$n equals 107$

Because of the higher $z$ -value for $99 percent sign$ confidence level

$n equals 98$

Because of the higher standard error at $95 percent sign$ confidence level

$n equals 107$

Because of the lower standard error at $95 percent sign$ confidence level

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