Problem 59 - Basic Statistics for Business & Economics

Question

The National Weight Control Registry tries to mine secrets of success from people who lost at least 30 pounds and kept it off for at least a year. It reports that out of 2,700 registrants, 459 were on a low-carbohydrate diet (less than 90 grams a day). a. Develop a \(95 \%\) confidence interval for the proportion of people on a low-carbohydrate diet. b. Is it possible that the population percentage is \(18 \% ?\) c. How large a sample is needed to estimate the proportion within \(0.5 \% ?\)

Step-by-Step Solution

Answer

a. CI: (0.1555, 0.1845); b. Yes, 18% is possible; c. Sample size needed: 21,883.

Step 1: Understand the Problem

We first identify the numbers involved in this exercise. We have 2,700 total registrants and 459 individuals on a low-carbohydrate diet. We need to find a 95% confidence interval for the proportion who are on this diet.

Step 2: Calculate Sample Proportion

Calculate the sample proportion \( p \) of people on a low-carbohydrate diet using: \[ p = \frac{459}{2700} \approx 0.17 \]

Step 3: Determine Standard Error

The standard error (SE) of the proportion is calculated using the formula: \[ SE = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.17 \times 0.83}{2700}} \approx 0.0074 \]

Step 4: Find the Z-value for 95% Confidence Interval

For a 95% confidence interval, the Z-value is 1.96. This is because we need 95% of the distribution, which leaves 2.5% in either tail of a normal distribution.

Step 5: Compute Confidence Interval

To find the 95% confidence interval, use: \[ CI = p \pm Z \times SE = 0.17 \pm 1.96 \times 0.0074 \] This gives us the interval \( (0.1555, 0.1845) \).

Step 6: Check Population Percentage Possibility

Determine if 18% (or 0.18) lies within our confidence interval. Since 0.18 is within \( (0.1555, 0.1845) \), it is possible.

Step 7: Calculate Sample Size for Desired Margin of Error

To find how large a sample is needed to estimate the proportion within 0.5% (or 0.005), use the formula: \[ n = \left( \frac{Z^2 \times p(1-p)}{E^2} \right) = \left( \frac{1.96^2 \times 0.17 \times 0.83}{0.005^2} \right) \] The calculated sample size \( n \approx 21,883 \).

Key Concepts

Understanding Sample Proportion

The concept of a sample proportion helps us estimate the proportion of people in a larger group, based on a smaller sample. In this exercise, we are interested in the proportion of people following a low-carbohydrate diet. To determine this, we take the number of people on such a diet (459), and divide it by the total number of registrants within our sample (2,700). This calculation yields a sample proportion denoted as \( p \): \[ p = \frac{459}{2700} \approx 0.17 \]Here, \( 0.17 \) implies that approximately 17% of our sample follows the diet. Sample proportions are useful in making predictions or decisions about entire populations from smaller, manageable groups.

Calculating Standard Error

Standard error measures the variability or spread of the sample proportion across different samples. It is fundamental in understanding how close our sample proportion is to the actual population proportion. The formula for the standard error \( SE \) of the sample proportion is:\[ SE = \sqrt{\frac{p(1-p)}{n}} \]For this particular case, with \( p = 0.17 \) and total sample size \( n = 2700 \):\[ SE = \sqrt{\frac{0.17 \times 0.83}{2700}} \approx 0.0074 \]This relatively small SE value indicates that the sample proportion is a precise estimate of the population proportion, thus boosting confidence in our inference.

Understanding Z-Value

The Z-value, or Z-score, is a statistical measure indicating how many standard deviations an element is from the mean. For a confidence interval, the Z-value helps determine how wide the interval will be. In the context of a 95% confidence interval, we utilize a Z-value of 1.96. This choice means that we are capturing 95% of the data under the normal distribution curve, leaving out 2.5% on either tail.The Z-value is crucial in the confidence interval formula, which is:\[ CI = p \pm Z \times SE \]In simple terms, higher confidence levels require larger Z-values, resulting in wider confidence intervals.

Sample Size Determination for Precision

Determining the right sample size is essential for achieving the desired precision in estimates. If we aim to have a margin of error within 0.5% for our estimate of a population proportion, we need to solve for sample size \( n \) using the formula:\[ n = \left( \frac{Z^2 \times p(1-p)}{E^2} \right) \]Where \( E \) is the desired margin of error (0.005 for 0.5%) and \( Z = 1.96 \) at a 95% confidence level. Inserting our values:\[ n = \left( \frac{1.96^2 \times 0.17 \times 0.83}{0.005^2} \right) \approx 21,883 \]This large sample size ensures our estimates are accurate within the specified margin. Calculations like these are pivotal when planning surveys or studies to ensure reliable results.