A bank wishes to estimate the mean balances owed by their MasterCard customers within $75. The population standard deviation is estimated to be $300. If a 98% confidence interval is used, how many cardholders should be sample.
Please provide sample size formula in calcutions.
The Roper Organization conducted identical surveys in 1990 and 2000. One question asked
women was, “Are most men basically kind, gentle, and thoughtful?” The 1990 survey revealed
that, of the 3,000 women surveyed, 2,010 said that they were. In 2000, 1,530 of the 3,000 women
surveyed thought that men were kind, gentle, and thoughtful. At the .05 level, can we conclude
that women think men are less kind, gentle, and thoughtful in 2000 compared with 1990?
To calculate the sample size needed to estimate the mean balances owed by MasterCard customers, we can use the formula:
n = (Z * σ / E)^2
where:
n = the required sample size
Z = the z-score corresponding to the desired confidence level
σ = the population standard deviation
E = the desired margin of error
In this case, the confidence level is 98%, which means the desired z-score corresponds to a cumulative probability of 0.99 in a standard normal distribution. Using a standard normal table, we can find that the z-score is approximately 2.33.
The population standard deviation (σ) is estimated to be $300, and the desired margin of error (E) is $75.
Substituting these values into the formula, we have:
n = (2.33 * 300 / 75)^2
Calculating this expression:
n = 9.32^2
n ≈ 86.8224
Since we cannot have a fractional number of customers, we'll round up to the nearest whole number.
Therefore, the bank should sample at least 87 cardholders to estimate the mean balances owed by MasterCard customers with a 98% confidence level and a margin of error of $75.