Question

Loan Amounts. B. Ciochetti et al. studied mortgage loans in the article "A Proportional Hazards Model of Commercial Mortgage Default with Originator Bias" (Joumal of Real Exfate and Economics, Vol, 27. No. 1. pp. 5-23). According to the article, the loan amounts of loans originated by a large insurance-company lender have a mean of $\backslash (6.74$ million with a standard deviation of $\backslash )15.37$ million. The variable "loan amount" is known to have a right-skewed distribution.

a. Using units of millions of dollars, determine the sampling distribution of the sample mean for samples of size $200$ . Interpret your result.

b. Repeat part (a) for samples of size $600$

c. Why can you still answer parts (a) and (b) when the distribution of loan amounts is not normal, but rather right skewed?

d. What is the probability that the sampling error made in estimating the population mean loan amount by the mean loan amount of a simple random sample of $200$ loans will be at most $\$1$ million?

e. Repeat part (d) for samples of size $600$

Short answer

Part (a) All sample means are normal, with a mean of $\$6.74$million and a standard deviation of $\$1.086823$million.

Part (b) All sample means are normal, with a mean of $\$6.74$million and a standard deviation of $\$0.627478$million.

Part (c) The loan amount distribution is not normal, yet the samples in (a) and (b) are big.

Part (d) The chance that the sampling error in forecasting the population mean loan amount from a sample of $200$ loans is at most $\$1$ million is $0.6424$

Part (e) The chance that the sampling error in forecasting the population mean loan amount from a sample of $200$loans is at most $\$1$million is $\$0.8882\$$

Step-by-step solution

Part (a) Step 1: Given information

The average loan amount created by a large insurance company lender is $6.74 million, with a standard variation of $15.37 million.

Part (a) Step 2: Concept

Formula used:$\text{population mean and standard deviation:}{\mu }_{\overline{x}}=\mu \text{and}{\sigma }_{\hat{x}}=\sigma /\sqrt{n}\text{.}$

Part (a) Step 3: Calculation

Let the population's average loan amount to be $\mu =\$6.74$

Let the loan amount population standard deviation be $\sigma =\$15.37$

Let the sample size be $n=200$

The sampling distribution of sample means' mean is,

$$\begin{gathered} {\mu }_x=\mu \\ =\$6.74 \end{gathered}$$

The sampling distribution of sample means' standard deviation is,

$$\begin{gathered} {\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}} \\ =\frac{15.37}{\sqrt{200}} \\ =\$1.086823 \end{gathered}$$

As a result, all sample means are normal, with a mean of $\$6.74$million and a standard deviation of $\$1.086823$million.

Part (b) Step 1:  Calculation

Let the population's average loan amount be$\mu =\$6.74$

Let the loan amount population standard deviation be $\sigma =\$15.37$

Let the sample size be $n=600$

The sampling distribution of sample means' mean is,

$$\begin{gathered} {\mu }_{\overline{x}}=\mu \\ =\$6.74 \end{gathered}$$

The sampling distribution of sample means' standard deviation is,

$$\begin{gathered} {\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}} \\ =\frac{15.37}{\sqrt{600}} \\ =\$0.627478 \end{gathered}$$

As a result, all sample means are normal, with a mean of $\$6.74$ million and a standard deviation of $\$0.627478$ million.

Part (c) Step 1: Calculation

The loan amount distribution is not normal, yet the samples in (a) and (b) are big. Regardless of the distribution of the variable under examination, the variable $\overline{x}$ is approximately normally distributed for a somewhat high sample size. With larger sample sizes, the approximation improves.

Part (d) Step 1: Calculation

Let $\bar{X}$ be the mean loan amount.

Find the value of $P(\mu-1 \leq \bar{X} \leq \mu+1)$.

Now,

$$\begin{gathered} P(\mu -1\le \overline{X}\le \mu +1)=P\left(\frac{\mu -1-\mu }{{\sigma }_{\overline{X}}}\le \frac{\overline{X}-\mu }{{\sigma }_{\overline{X}}}\le \frac{\mu +1-\mu }{{\sigma }_{\overline{X}}}\right) \\ =P\left(-\frac{1}{1.086823}\le z\le \frac{1}{1.086823}\right) \\ =P(-0.92\le z\le 0.92) \\ =P(z\le 0.92)-P(z\le -0.92) \\ =0.8212-0.1788 \\ =0.6424 \end{gathered}$$

As a result, the chance that the sampling error in forecasting the population mean loan amount from a sample of $200$ loans is at most $\$1$ million is $0.6424$

Part (e) Step 1: Calculation

Repeat part (d) with sample of size is $600$

Let $x$is the loan amount.

Find the value of $P(\mu -1\le \overline{X}\le \mu +1)$

Now,

$$\begin{gathered} P(\mu -1\le \overline{X}\le \mu +1)=P\left(\frac{\mu -1-\mu }{{\sigma }_{\overline{X}}}\le \frac{\overline{X}-\mu }{{\sigma }_{\overline{X}}}\le \frac{\mu +1-\mu }{{\sigma }_{\overline{X}}}\right) \\ =P\left(-\frac{1}{0.627478}\le z\le \frac{1}{0.627478}\right) \\ =P(-1.59\le z\le 1.59) \\ =P(z\le 1.59)-P(z\le -1.59) \\ =0.9441-0.0599 \\ =0.8882 \end{gathered}$$

As a result, the chance that the sampling error in forecasting the population mean loan amount from a sample of $200$ loans is at most $\$1$ million is $0.8882$