Question

2. Market researchers want to estimate the mean monthly expenditure on transport for households in Sydney. They randomly selected 19 householders and asked their expenditure on transport, in dollars, over the past month. Data from the selected households are given below. 181 266 339 292 527 150 484 488 282 571 536 153 583 434 135 187 497 144 520

Part A

What is the best point estimate of the population mean, to two decimal places? Give your answer in the form xxx.xx Data can be copied and pasted into Excel to complete the calculations. Answer:

Part B Now assume that the population standard deviation is known to be 170. To two decimal places, what is the standard error for the mean of the sample? Give your answer in the form xx.xx Answer:

Part C Based on the given sample and the known standard deviation, calculate a 90% confidence interval for mean monthly expenditure on transport. What are the lower and upper limits of the confidence interval? Give your answers to one decimal place, in the form xxx.x

Lower limit: Answer

Upper limit: Answer

Part D True or false: A 99% confidence interval based on the same sample would be narrower. To answer, type T or F in the space provided. Answer:

Answer 1

Part A:

From the given data, Sample mean () is calculated as follows:

356.26

Part B:

Standard Error (SE) is given by:

So,

39.00

Part C:

= 0.10

From Table, critical values of Z = 1.645

Lower Limit = 356.26 - (1.645 X 39.00) = 356.26 - 64.156 = 292.10

Upper Limit = 356.26 + (1.645 X 39.00) = 356.26 + 64.156 = 420.42

Part D:

A 99% confidence interval would be wider.

So,

F