The data include average annual household income levels of citizens of selected

ID: 3183598 • Letter: T

Question

The data include average annual household income levels of citizens of selected U.S. cities.

A. Use Excel’s StatTools to obtain a simple random sample of size 10 from this frame.

B. Using the sample generated in Part A, construct a 95% confidence interval for the mean average annual household income level of citizens in the selected U.S. cities. Assume that the population consists of all average annual household income levels in the given frame.

C. Interpret the 95% confidence interval constructed in Part B.

D. Does the 95% confidence interval contain the actual population mean? If not, explain why not. What proportion of many similarly constructed confidence intervals should include the true population mean value?

City Index Household Income 1 $54,300 2 $61,800 3 $61,400 4 $50,800 5 $56,200 6 $48,300 7 $61,600 8 $63,200 9 $55,200 10 $58,000 11 $77,600 12 $47,600 13 $62,700 14 $46,200 15 $64,300 16 $56,000 17 $53,400 18 $56,800 19 $51,200 20 $59,000 21 $53,500 22 $45,600 23 $70,100 24 $108,700 25 $46,400 26 $56,700 27 $59,100 28 $46,300 29 $52,900 30 $56,300 31 $67,300 32 $63,800 33 $70,600 34 $49,800 35 $51,300 36 $56,600 37 $49,600 38 $67,400 39 $53,700 40 $48,700 41 $61,500 42 $53,000 43 $51,000 44 $55,600 45 $51,600 46 $57,200 47 $54,300 48 $51,500 49 $53,500 50 $61,800 51 $44,800 52 $57,400 53 $48,100 54 $52,700 55 $57,400 56 $65,500 57 $59,600 58 $62,000 59 $49,700 60 $54,400

Explanation / Answer

Part a

The simple random sample of size 10 by using excel is given as below:

(Data>Data Analysis>Random Number Generation or =randbetween command)

City Index

Household Income

$77,600

$49,600

$46,400

$51,600

$53,400

$61,400

$52,700

$54,400

$57,200

$56,000

Part b

Confidence interval = Xbar -/+ t*SD/sqrt(n)

Standard error = SD/sqrt(n)

For the given data, we have

Sample Standard Deviation

8627.095559

Sample Mean

$56,030

Sample Size

Confidence Level

95%

Standard error = 8627.1/sqrt(10) = 2728.13

t-value = 2.2622

Confidence interval = 56030 -/+ 2.2622*2728.13

Confidence interval = 56030 -/+ 6171.4524

Lower limit = 56030 – 6171.4524 = 49858.55

Upper limit = 56030 + 6171.4524 = 62201.45

Part c

We are 95% confident that the population annual household income will be lies between two values $49858.55 and $62201.45.

Part d

Yes, the given confidence interval contain the actual population mean $57043. About 95% confidence intervals will contain true population mean value.