A survey of community banks asked about the loan to deposit ratio (LTDR) a banks total loans as a percent of its total deposits. The mean LTDR for the 110 banks in the sample is xbar = 76.7 and the standard deviation is s = 12.3. In the text a 95% confidence interval (using a z score) is given for the mean LTDR for community banks. This solution is theoretically not completely correct. What should the interval be?
A -
74.6-79.2
74.38-79.02
74.3-78.9
74.28-78.88
74.2-78.8
Formula:
CI95 = mean + or - 1.96(sd divided by √n)
...where + or - 1.96 represents the 95% confidence interval using a z-table, sd = standard deviation, √ = square root, and n = sample size.
With your data:
CI95 = 76.7 + or - 1.96(12.3/√110)
Finish the calculation for your interval.
To calculate the interval, we first need to determine the value of 1.96 from the z-table. The value corresponds to a 95% confidence level, which means that the area under the normal distribution curve outside the interval is 0.05 (5%).
Using the z-table, we find that the value of 1.96 corresponds to the area of 0.975. Since the area under the normal distribution curve is symmetric, we subtract 0.975 from 1 to get the area in the left tail, which is 0.025.
Next, we calculate the standard error (SE) using the formula:
SE = standard deviation / √(sample size)
SE = 12.3 / √110
Now, we can calculate the margin of error (ME) by multiplying the SE by 1.96:
ME = 1.96 * SE
Finally, we can calculate the confidence interval (CI) by adding and subtracting the margin of error from the mean:
CI95 = x̄ ± ME
Substituting in the values we have:
CI95 = 76.7 ± ME
CI95 = 76.7 ± (1.96 * (12.3 / √110))
Calculate the value of ME and substitute it back into the equation to get the final interval.