1) Mobile and Exxon, the nation s\' two largest oil companies have recently redu
ID: 3255656 • Letter: 1
Question
1)
Mobile and Exxon, the nation s' two largest oil companies have recently reduced their wholesale prices for gasoline. Mobile claims that their gas stations now sell, on the average, one gallon of regular gas at a lower price than Exxon brand. In a recent study the price of one gallon of regular gas for 8 randomly selected Mobile gas stations has a mean of $2.20 and a standard deviation of $0.11; and the price of one gallon of regular gas in 12 randomly selected Exxon gas stations have a mean of $2.29 and a standard deviation of $0.07. Assuming that the two population standard deviations are equal, a 99% confidence interval for the true difference in average price per one gallon of regular gas of Mobile brand and Exxon brand of gas is given by:
a. (2.20-2.29)+-2.878 ((.11)^2 /8 + (.07)^2 /12)^.5
b. (2.20-2.29)+-2.528 {[(.11)^2 / 8] + [(.07)^2 / 12]}^.5
c. (2.20-2.29)+-2.878 {[[7(.11)^2 + 11(.07)^2] /18][(1/8) + (1/12)]}^.5
d. (2.20-2.29)+-2.552 {[[7(.11) + 11(.07)] /18][(1/8) + (1/12)]}^.5 2)
2) Suppose two independent random samples of size n1=5 and n2=9 are taken from two independent normal populations, respectively. The computed sample standard deviations are s1=4 and s2=7, respectively. A 98% confidence interval for the ratio, R=(first population standard deviation / second population standard deviation), is:
a. [0.4549, 4.6339]
b. [1.2198 , 2.1912]
c. [0.2158, 2.1983]
d. [0.2069, 21.4592]
Explanation / Answer
According to chegg policy ,I am going to answer the first one
1 - Mobile gas stations has a mean of $2.20 and a standard deviation of $0.11;
2 - Exxon gas stations have a mean of $2.29 and a standard deviation of $0.07
n1 = 8 ,n2 = 12
df = n1 + n2 - 2 = 18
t 0.005,18 = 2.878
sp = sqrt(s1^2 *(n1-1) + s2^2 *(n2 -1)) /(n1 +n2 - 2))
= sqrt( (7 *0.11^2 + 11 * 0.07^2 )/18 )
=0.087749
confidence interval is
(X1bar -x2bar ) + - t * sp sqrt(1/n1 +1 /n2)
hence option
c. (2.20-2.29)+-2.878 {[[7(.11)^2 + 11(.07)^2] /18][(1/8) + (1/12)]}^.5 is correct
hence
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