Question

149 0.8175 Suppose you analyzed the average monthly credit card bill of Visa credit card custome cand customers Sample ri and one with 109 Visa credit card customers Sample #2 Every credit card customer in Sample #1 had a monthly credit card bill of $1,809. Sample #2, on the other hand, had an average monthly credit card bill of $1,879 with a variance of 1,361,788 (S)A2. If you combined these samples into one... Note: Answer Questions 5-8 by Using the Excel Output The U.S. Department of Energy's Fuel Economy Guide provides fuel efficiency data for cars and trucks. (See exercise 35 on page 77 in your textbook for a 1similar problem.) Use the data in the FuelData sheet in this workbook to generate excel output to answer the following questions. Please note that units 2 are "liters for engine size and "miles per gallon" for highway fuel efficiency. Put the output in given space on FuelData sheet. Each orange numerical MUST

Answer 1

Solution

Back-up Theory

Properties of Mean

If xi is the same for all i = 1 to n, say x, then mean = nx……………………………...........................................…............….. (1)

If Xbar is the mean of X based on n₁ sample observations and Ybar is the mean of Y based on n₂ sample observations,

then the combined mean, say M = (n₁Xbar + n₂Ybar)/(n₁ + n₂) ….....................................................................................…. (2)

Properties of Variance

If xi is the same for all i = 1 to n, say x, then variance = 0 …………….....................................................…………………….. (3)

If s₁² is the variance of X based on n₁ sample observations on X and s₂² is the variance of Y based on n₂ sample observations on Y, then the combined variance, say s²= {(n₁- 1)s₁² + (n₂ - 1)s₂²}/(n₁ + n₂ - 2) ………………………….…. (4)

Now to work out the solution,

Given n₁ = 149, Xbar = 1809 [vide (1)], s₁² = 0 [vide (3)], n₂ = 109, Ybar = 1879, s₂² = 1361788,

Part (a)

Vide (2), combined average (mean) = {(149 x 1809) + (109 x 1879)}/(149 + 109)

= $1838.6   Answer

Part (b)

In the combined sample, there are 258 values. So, the combined sample median would be the average of 129^th and 130^th values in the ordered set of the combined values.

Case 1 No value in the second sample is less than 1809. Then, the ordered set would be:

1809, 1809, ….., 1809, (149 values), followed by 109 values of the second sample and hence both 129^th and 130^th values in the ordered set of the combined values would be 1809. Thus, the median = 1809

Case 2 m values in the second sample are less than 1809.

Since mean of the second sample is 1879, at least one value must be greater than 1879 and hence greater than 1809. So, m < 109, say 108 as an extreme case. Then, the ordered set would be:   

108 values of the second sample, 1809, 1809, ….., 1809, (149 values), followed by 1 value of the second sample and hence both 129^th and 130^th values in the ordered set of the combined values would be 1809. Thus, the median = 1809.

Consequently, the median of the combined sample values is always 1809 Answer

Part (c)

Since, as brought out in the above analyses, median > mean the distribution would be

positively skewed, i.e., longer right tail. Answer

Part (d)

Vide (4), combined variance = {(148 x 0) + (108 x 1361788}/(148 + 108)

= 574504

So, combined standard deviation = sqrt(574504)

= $756.96 Answer

DONE