A credit manager has chosen a sample of size 2,164 to estimate the mean balance of credit accounts to within $0.50. A pilot study of 50 accounts had a mean of $1,219 and a variance of 100. The level of confidence of the interval is:
90%
95%
98%
99%
To determine the level of confidence of the interval, we need to calculate the margin of error and then use it to find the corresponding confidence level.
First, we need to find the standard deviation (σ) of the population. Since we don't have the population standard deviation, we can use the sample variance (s^2) from the pilot study to estimate it. The sample variance is given as 100. To get the standard deviation, we take the square root of the variance: σ = √100 = 10.
Next, we calculate the standard error of the mean (SE) based on the sample size (n) and the standard deviation (σ). The standard error of the mean is given by the formula: SE = σ / √n. In this case, n = 2,164, so the standard error is SE = 10 / √2164 ≈ 0.214.
The margin of error (ME) is determined by multiplying the standard error by the appropriate critical value from the t-distribution. The critical value is chosen based on the desired level of confidence. Common choices are 90%, 95%, 98%, and 99%.
For a 90% confidence level, the critical value can be obtained from the t-distribution table or a statistical calculator. Assuming a two-tailed test (since we want to estimate the mean balance without a specific direction), the critical value is 1.645.
The margin of error is then calculated as ME = critical value * standard error = 1.645 * 0.214 ≈ 0.352.
Since we want to estimate the mean balance to within $0.50, which is the margin of error, we need to find the corresponding level of confidence where the margin of error is equal to $0.50.
Therefore, we need to compare the margin of error (ME) with the desired interval of $0.50.
If ME < $0.50, then the level of confidence is higher than the desired interval. We can proceed to check the remaining options.
If ME > $0.50, then the level of confidence is lower than the desired interval. We eliminate this option.
In this case, ME ≈ 0.352, which is greater than $0.50. Therefore, the level of confidence for the interval is less than or equal to 90%.
So, the answer is: 90%.