Question

A bank manager wants to know the mean amount of mortgage paid per month by homeowners in an area. A random sample of homeowners selected from this area showed that they pay an average of per month for their mortgages. The population standard deviation of such mortgages is . a. Find a confidence interval for the mean amount of mortgage paid per month by all homeowners in this area. Round your answers to two decimal places. 1538.51 to 1627.49 dollars b. Suppose the confidence interval obtained in part a is too wide. Select all of the ways the width of this interval can be reduced.

Answer 1

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

The 97% confidence interval is
`1528  <  \mu <  1612`

b

Option A and Option D are correct

Explanation:

From the question we are told that

The sample size is n = 119

The sample mean is
`\$ 1570`

The population standard deviation is
`\sigma  =  \$ 211`

Generally given that the confidence level is 97% then the level of significance is mathematically represented as

`\alpha = (100 - 97 )\%`

=>
`\alpha = 0.03`

From the normal distribution table the critical value of
`(\alpha )/(2)` is

`Z_{(\alpha )/(2) } =  2.17`

Generally the margin of error is mathematically represented as

`E =  Z_{(\alpha )/(2) } *  (\sigma)/(√(n) )`

`E =  2.17*  (211)/(√( 119 ) )`

=>
`E =   41.98`

Generally the 97% confidence interval is mathematically represented as

`\= x  - E  <  \mu <  \= x  + E`

=>
`1570  - 41.98  <  \mu <  1570  + 41.98`

=>
`1528  <  \mu <  1612`

Generally looking at confidence interval we see that the width is dependent on the size of the margin of error.

Generally the margin error is directly proportional to the critical value of
`(\alpha )/(2)` which increases when the confidence level increases and vise versa

Also the margin error is inversely proportional to the square root of sample size

Hence to reduce the width of the confidence interval we need to lower the confidence level and increase the sample size