Question

A bank is experimenting with programs to direct bill companies for commercial loans. They are particularly interested in the number of errors of a billing program. To examine a particular program, a simulation of 1000 typical loans is run through the program. The simulation yielded a mean of 4.6 errors with a standard deviation of 0.5. Consturct a 95% confidence interval on the true mean error rate.

Answer 1

Answer:

95% confidence interval on the true mean error rate = [4.57 , 4.63] .

Step-by-step explanation:

We are given that to examine a particular program, a simulation of 1000 typical loans is run through the program.                                                         The simulation yielded a Mean,$Xbar$ = 4.6 errors with a Standard deviation,s = 0.5.

Since, here we know nothing about population standard deviation so we will use t statistics quantity here i.e.;

                       $\frac{Xbar-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$  where, $Xbar$ = sample mean

                                                              s    = sample standard deviation

                                                              n  = sample size(no. of simulations)

                                                              $\mu$ = population mean or true mean

So, 95% confidence interval on the true mean error rate is given by;

P(-1.96 < $t_9_9_9$ < 1.96) = 0.95 {because at 5% significance level t table gives

                                                  value close to 1.96}

P(-1.96 < $\frac{Xbar-\mu}{\frac{s}{\sqrt{n} } }$  < 1.96) = 0.95

P(-1.96 * $\frac{s}{\sqrt{n} }$ < $Xbar - \mu$ < 1.96 * $\frac{s}{\sqrt{n} }$ ) = 0.95

P(-Xbar - 1.96 * $\frac{s}{\sqrt{n} }$ < $-\mu$ < Xbar - 1.96 * $\frac{s}{\sqrt{n} }$ ) = 0.95

P( Xbar - 1.96 * $\frac{s}{\sqrt{n} }$ < $\mu$ < Xbar + 1.96 * $\frac{s}{\sqrt{n} }$ ) = 0.95

95% Confidence interval for $\mu$ = [Xbar - 1.96 * $\frac{s}{\sqrt{n} }$ , Xbar + 1.96 * $\frac{s}{\sqrt{n} }$ ]

                                                  = $[4.6 - 1.96*\frac{0.5}{\sqrt{1000} } , 4.6 + 1.96*\frac{0.5}{\sqrt{1000} } ]$

                                                  = [4.57 , 4.63]

Therefore, 95% confidence interval on the true mean error rate is        [4.57 , 4.63] .

Answer 2

Answer:

The 95% confidence interval for true mean error is (4.57, 4.63).

Step-by-step explanation:

Let X = the number of errors of a billing program.

According to the Central limit theorem if a large sample (n > 30) is drawn from an unknown population then the sampling distribution of the sample mean will follow a Normal distribution with mean ($\mu_{\bar x}=\mu$) and standard deviation      ($\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}$).

The sample size of the loans is, n = 1000.

The mean is, ${\bar x}=4.6$.

The standard deviation is,  $\sigma_{\bar x}=\frac{0.50}{\sqrt{1000}}=0.016$

The confidence interval for mean is:

$CI=\bar x\pm z_{\alpha /2}\times \sigma_{\bar x}$

The critical value of z for 95% confidence interval is:

$z_{\alpha /2}=z_{0.05/2}=z_{0.025}=1.96$

**Use the z-table for critical values.

Compute the 95% confidence interval for true mean error as follows:

$CI=\bar x\pm z_{\alpha /2}\times \sigma_{\bar x}\\=4.6\pm1.96\times0.016\\=4.6\pm0.03136\\=(4.56864, 4.63136)\\\approx(4.57, 4.63)$

Thus, the 95% confidence interval for true mean error is (4.57, 4.63).