3) You are working for a government department and your boss, Jane, has asked yo

Question

3) You are working for a government department and your boss, Jane, has asked you to calculate some results on weekly household income across the state, including a 95% confidence interval for the mean weekly household income that she needs to include in a report. She also says that she is not sure exactly what a 95% confidence interval means and would like you to add an explanation.

You have been supplied with a sample of weekly income figures for 110 households. The data is presented here:

Download the data

Historically, the standard deviation in weekly household income is $458.

Complete the report to your boss. Give your numeric answers to 2 decimal places.

Dear Jane,

Here are the results gathered from the collected data:

Assuming a population standard deviation in weekly household income of $458, the 95% confidence interval for the mean weekly household income is:

a)? ?

b)This means that :

a. approximately 95% of sample means will be within the interval given above
b. on approximately 95% of days in a given period the stock makes a return within the interval given above
c. the population mean weekly household income is definitely within the interval given above
d. using a process that gives correct results in 95% of cases, the population mean weekly household income is within the interval given above

Weekly household income ($) 1707 1917 2221 364 2162 2029 2222 1699 910 1843 2186 1998 722 1519 1817 1642 1475 1201 1688 1458 1531 1235 919 1781 1881 2216 1239 1185 1484 969 1517 1690 1951 1555 1756 2101 812 2170 1915 1405 1295 1790 1839 1601 1765 1061 919 1742 1814 1845 1225 2109 2313 1523 1604 1642 1816 1121 1386 818 1481 1329 1351 2398 1027 2011 1720 2078 1648 1462 890 2380 1504 1896 2270 1989 1406 954 2238 1474 1567 942 2151 766 2312 1735 1495 1423 1193 1302 2031 2380 2359 957 1854 1770 2265 1436 1191 1688 1552 1291 1910 1406 2127 2244 2017 1630 2062 2209

Explanation / Answer

xbar= sum/n= 180091/110= 1637.191

since n> 30 we will use Z statistic to compute this 95% confidence interval

M = 1637.191
t = 1.96
sM = (4582/110) = 43.67

= M ± Z(sM)
= 1637.191 ± 1.96*43.67
= 1637.191 ± 85.58886

M = 1637.191, 95% CI [1551.60214, 1722.77986].

You can be 95% confident that the population mean () falls between 1551.60214 and 1722.77986.

This means that : approximately 95% of sample means will be within the interval given above.

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