Question

An economist is interested in studying the monthly incomes of consumers in a particular region. The population standard deviation of monthly income is known to be 1000 rupees. A random sample of 49 individuals resulted in an average monthly income of 15000 rupees. What is the total width of the 90% confidence interval? *

Answer 1

The total width of the 90% confidence interval is 469.972.

Step-by-step explanation:

We are given that an economist is interested in studying the monthly incomes of consumers in a particular region. The population standard deviation of monthly income is known to be 1000 rupees.

A random sample of 49 individuals resulted in an average monthly income of 15000 rupees.

Firstly, the pivotal quantity for 90% confidence interval for the population mean is given by;

                              P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$  ~ N(0,1)

where, $\bar X$ = sample average monthly income = Rs 15,000

            n = sample of individuals = 49

            $\sigma$ = population standard deviation = Rs 1,000

            $\mu$ = population mean monthly income

Here for constructing 90% confidence interval we have used One-sample z statistics because we know about the population standard deviation.

So, 99% confidence interval for the population mean, $\mu$ is ;

P(-1.6449 < N(0,1) < 1.6449) = 0.90  {As the critical value of z at 5%

                                     level of significance are -1.6449 & 1.6449}

P(-1.6449 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.6449) = 0.90

P( $-1.6449 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.6449 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.6449 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.6449 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X-1.6449 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.6449 \times {\frac{\sigma}{\sqrt{n} } }$]

                                        = [ $15000-1.6449 \times {\frac{1000}{\sqrt{49} } }$ , $15000+1.6449 \times {\frac{1000}{\sqrt{49} } }$ ]

                                        = [Rs 14765.014 , Rs 15234.986]

Therefore, 99% confidence interval for the population mean monthly income is [Rs 14765.014 , Rs 15234.986].

Now, the total width of 90% confidence interval = Rs 15234.986 - Rs 14765.014 = Rs 469.972.