A well-known bank credit card firm wishes to estimate the proportion of credit card holders who carry a nonzero balance at the end of the month and incur an interest charge. Assume that the desired margin of error is .03 at \(98 \%\) confidence. a. How large a sample should be selected if it is anticipated that roughly \(70 \%\) of the firm's card holders carry a nonzero balance at the end of the month? b. How large a sample should be selected if no planning value for the proportion could be specified?

Short Answer

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a) A sample size of 765 should be selected for an anticipated proportion of 70% credit card holders carrying a nonzero balance at the end of the month, with a desired margin of error of 0.03 at 98% confidence. b) A sample size of 1068 should be selected when no planning value for the proportion is specified, using the same desired margin of error of 0.03 at 98% confidence.

Step by step solution

01

a) Sample size estimation with an anticipated proportion of 70%

In this case, we are given an anticipated proportion (p) of 70%, a margin of error (E) of 0.03, and a confidence level of 98%. 1. Find the z-score (Z) corresponding to the confidence level of 98%. A 98% confidence level corresponds to a 1% tail on each side (100% - 98% = 2%, divided equally into left and right tails). We need to find the Z value for which there is 99% (98% + 1%) of the area under the standard normal distribution. The Z value for this case is approximately 2.33. 2. Apply the formula for sample size estimation for proportions: \(n = \frac{Z^2 * p * (1-p)}{E^2}\) Where Z = 2.33, p = 0.7, and E = 0.03. 3. Calculate the result: \(n = \frac{(2.33)^2 * 0.7 * (1-0.7)}{(0.03)^2}\) \(n ≈ 764.73\) 4. Round up the sample size: Since the sample size must be an integer, we round it up to the nearest whole number: n = 765. Therefore, a sample size of 765 should be selected to fulfill the requirements in scenario a.
02

b) Sample size estimation with no planning value for the proportion

When there is no planning value specified for the proportion, we assume the worst-case scenario, i.e., the most conservative assumption for the proportion, to be p = 0.5 or 50%. This is because the variability of p(1-p) is maximum when p = 0.5. 1. Use the same z-score as calculated in scenario a, Z = 2.33, a margin of error (E) of 0.03, and a new conservative proportion p = 0.5. 2. Apply the formula for sample size estimation for proportions: \(n = \frac{Z^2 * p * (1-p)}{E^2}\) Where Z = 2.33, p = 0.5, and E = 0.03. 3. Calculate the result: \(n = \frac{(2.33)^2 * 0.5 * (1-0.5)}{(0.03)^2}\) \(n ≈ 1067.07\) 4. Round up the sample size: Since the sample size must be an integer, we round it up to the nearest whole number: n = 1068. Therefore, a sample size of 1068 should be selected in scenario b, where no planning value for the proportion was specified.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Confidence Interval
A confidence interval is a range of values, derived from sample statistics, which is likely to contain the value of an unknown population parameter. It is used as a tool to express the reliability and precision of an estimate. Specifically in our exercise, a 98% confidence interval suggests that we can be 98% certain that the true proportion of credit card holders who carry a nonzero balance at the end of the month falls within our calculated range. The confidence interval balances precision with certainty: the wider the interval, the more certain we can be that the population parameter lies within it, but the less precise is our estimate.

Confidence intervals are constructed using the point estimate, such as the sample mean or proportion, and a margin of error, which contributes to determining the range's width. The margin of error itself is influenced by the size of the sample and the level of confidence desired. A common misconception is that a 98% confidence interval is a prediction interval that will contain the true value 98% of the time in repeated samples; instead, it reflects the interval calculation method, with 98% confidence in the process, not the outcome.
Margin of Error
The margin of error represents the degree of error to be expected in the sampling process and reflects the maximum expected difference between the true population parameter and a sample estimate. In our case, the margin of error of 0.03 indicates that the true proportion of credit card holders with nonzero balances could be up to 3 percentage points higher or lower than our sample estimate. The margin of error is directly influenced by several factors:
  • The size of the sample: larger samples have a smaller margin of error.
  • The variability in the data: more variability implies a greater margin of error.
  • The desired level of confidence: higher confidence levels lead to a larger margin of error.
Understanding the margin of error is crucial for interpreting survey results because it provides a context for how much the survey results might differ from the actual population value.
Proportion
A proportion in a statistical sense refers to the fraction or percentage of occurrences or characteristics within a dataset. In the provided exercise, the term proportion relates to the percentage of credit card holders who carry a nonzero balance at the end of the month. Estimated at 70% (or 0.7) for the first scenario in the exercise, the proportion helps us to calculate the sample size needed to achieve the desired margin of error with a given confidence interval. Notably, when there is no preliminary data to estimate the proportion, the most conservative estimate of 0.5 (or 50%) is used because it assumes maximal uncertainty and variability, which in turn affects the minimum sample size required for the survey. This ensures the sample is robust enough to make a reliable inference about the population, regardless of the actual proportion.
Z-Score
A z-score, also known as a standard score, quantifies the number of standard deviations an element is from the mean of its distribution. For confidence intervals, the z-score is associated with the desired level of confidence and represents the buffer for error we are willing to accept. In the context of the exercise, a z-score of approximately 2.33 is used because it corresponds to the tails (1% each) of the standard normal distribution at a 98% confidence level. This score is crucial because it scales the margin of error according to the confidence level chosen. The higher the desired confidence level, the higher the z-score and the wider the confidence interval will have to be to accommodate that level of certainty. Thus, the z-score plays an essential role in determining the sample size necessary to estimate the population parameter (in this case, the proportion) within the specified margin of error.

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