Question 1 Homework 5 Due October 16, 2020 By 11:59 Pm

Question

Question 1homework 5 Due October 16, 2020 by 1159 Pmquestion Onea W Question one: 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 70% of the firm's cardholders 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?

Dr. Jack HW Helper · Accepted Answer

The task involves determining the appropriate sample size for estimating the proportion of credit card holders who carry a nonzero balance at the end of the month, with specific confidence levels and margins of error. The foundational concept is to use the formula for sample size calculation in proportion estimates. This calculation varies depending on prior knowledge of the proportion involved. For part (a), the key is to calculate the sample size when an anticipated proportion $ p $ is known, here estimated as 0.70. The formula for the required sample size $ n $ to estimate a population proportion with a given margin of error $ E $ and confidence level is: $$ n = \frac{Z^2 \times p \times (1 - p)}{E^2} $$ Where: - $ Z $ is the z-score corresponding to the confidence level - $ p $ is the estimated proportion - $ E $ is the margin of error At a 98% confidence level, the critical z-score $ Z $ is approximately 2.33, derived from standard normal distribution tables. The desired margin of error is 0.03. Plugging the values into the formula: $$ n = \frac{(2.33)^2 \times 0.70 \times 0.30}{0.03^2} $$ $$ n = \frac{5.4289 \times 0.21}{0.0009} $$ $$ n = \frac{1.139 \, \text{(approximate numerator)}}{0.0009} $$ $$ n \approx 1265 $$ Thus, a sample size of approximately 1,265 cardholders is necessary to estimate the proportion with the specified precision. In part (b), where no prior estimate of the proportion is available, the conservative approach is to assume the maximum variability, i.e., $ p = 0.5 $. Using this value ensures the sample size is sufficiently large regardless of the actual proportion. Reapplying the formula: $$ n = \frac{(2.33)^2 \times 0.5 \times 0.5}{0.03^2} $$ $$ n = \frac{5.4289 \times 0.25}{0.0009} $$ $$ n = \frac{1.3572}{0.0009} $$ $$ n \approx 1508 $$ Therefore, approximately 1,508 observations are required if no prior estimate exists. These calculations are crucial for planning surveys and ensuring that the proportion is estimated accu

Question 1homework 5 Due October 16, 2020 by 1159 Pmquestion Onea W

Paper For Above instruction

Related Assignments