Is Hit A

STATISTICS ASSIGNMENTConfidence Interval

(Two populations)

6 Problems

By: Ishita ChawraDate:- 27th Aug,2010.

Q.1. Until a few years ago, the market for consumer credit was considered to be segmented. Higher income, higher spending people tended to be American Express cardholders, and lower income, lower spending people were usually Visa cardholders. In the last few years, Visa has intensified its efforts to break into the higher income segments of the market by using magazine and TV advt. to create a high class image. Recently, a consulting firm was hired by Visa to determine whether avg. monthly charges on the American express Gold card are approximately equal to the average monthly charges on preferred Visa. A random sample of 1,200 preferred Visa cardholders was selected, and the sample average monthly charge was found to be x1 = $452. An independent random sample of 800 Gold card members revealed a sample mean x2 = $523. Assume SD1 = $212 and SD2 = $185. (Holders of both the Gold card and preferred Visa were excluded from the study.) Can you establish a 95% confidence interval for the difference between the average monthly charge on the American Express Gold Card and the average monthly charge on the Preferred Visa Card?

Ho: µ1- µ2 = 0

H1: µ1- µ2 ≠ 0

Formula:-

x1- x2 ± z /2

523 – 452 ± 1.96 √ 2122 /1200 + 1852 /800

71 ± 1.96 √80.23

71 ± 1.96 * 8.957

53.44, 88.56 Ans…

*z /2 = with the diagram = 1-α = 1- .95

α= 0.05, /2 = 0.025 (Significance region)

Confidence region = 0.475 i.e. 1.96 through the statistics table.

CONCLUSION ----The consulting firm may report to Visa that it is 95% confident that the average American Express Gold Card monthly bill is anywhere from $53.44 to $88.56 higher than the average Preferred Visa bill.

Q.2. Suppose that the makers of Duracell batteries want to demonstrate that their size AA

battery lasts an average of at least 45minutes longer than Energizer. Two independent random

samples 100 batteries of each kind are selected, and the batteries are run continuously until

they are no longer operational. The sample average life for Duracell is found to be X1= 308

minutes. The result for the energizer battery is X2 = 254 minutes. Assume SD1 = 84minutes and

SD2 = 67minutes. Can we substantiate with 95% confidence interval that Duracell’s batteries

last, on average 45minutes longer than Energizer batteries of the same size?

Ans….

Ho: µ1- µ2 ≤ 45

H1: µ1- µ2 > 45

Formula:-

x1- x2 ± z /2

308-254 ± 1.96 √ (842 / 100 + 672 /100)

54 ± 1.96 √ (70.56+ 44.89)

54 ± 1.96 * 10.74 = (32.95, 75.05) Ans …




Thus, we conclude that with 95% confidence level we can find the range in which

the battery will last i.e. (32.95, 75.05)

* The z value when calculated stated 0.838 which lies in no rejection area of Ho. Thus, there is

no sufficient evidence to support Duracell’s batteries claim.

Q.3 LINC is a software tool developed by Burroughs Corporation. The program automatically

writes some of the coding that programmers have to do manually. LINC supposedly saves

programming time and allows programmers to operate more efficiently. In a test of the

software package, 45 programmers (group1) were asked to write a program without LINC and

then run the program until it performed with no bugs. The times from start to finish for this

group were recorded. Group 2 consisted of 32 programmers who were asked to prepare the

same program with the aid of LINC. Before getting the data, the protocol established was to run

the test to prove that the package reduces the avg. programming time. The results wereX1 = 26

minutes, X2 = 21 minutes, s1 = 8 minutes and s2 = 6 minutes. Conduct the test, and state your

conclusion. Is LINC effective in reducing average programming time, conduct a 90% confidence

interval.

Formula:-

x1- x2 ± z /2

26-21 ± 1.65 √82 /45 + 62 /32

5 ± 1.65 √2.545

5 ± 1.65 * 1.59

2.386, 7.632 Ans…




Conclusion --- with 90% Confidence we can say that, the programming time lies between

2.386min to 7.632min.

Q.4 The manufacturers of compact disk players want to test whether a small price reduction

is enough to increase sales of their product. Randomly chosen data on 15 weekly sales totals at

outlets in a given area before the price reduction show a sample mean of $ 6,598 and a sample

SD of $844. A random sample of 12 weekly sales totals after the small price reduction gives a

sample mean of $6, 870 and a sample SD of $669. Is there evidence that the small price

reduction is enough to increase sales of compact disk players?

Answer:-

We can conduct a confidence interval (CI) for the parameter in question—here, the difference

between the two populations means. The CI for this parameter is based on the t distribution

with n1 + n2 – 2 degrees of freedom. The 95% CI for µ1 - µ2 is ----

Ho: µ1- µ2 ≥ 0

H1: µ1- µ2 < 0

= n1 + n2 – 2

= 15+ 12-2

= 25 degree of freedom.

Formula:-

x1- x2 ± t0.025 √Sp2 (1 / n1 + 1/ n2)

(6870- 6598) ± 2.06 √(0.15)(595,835)

272 ± 2.06 * 298.96

-343.85 ,887.85

We see that the confidence interval indeed contains the null hypothesized difference of zero. Since, the test resulted in non rejection of the null hypothesis; the CI contained the null hypothesis difference between the two population means: zero. This is due to the connection between hypothesis test and Confidence interval.

Q.5 From time to time, Bank America Corporation comes out with its free and easy travelers Cheques sweepstakes, designed to increase the amounts of Bank America traveler’s checks sold. Since the amount bought per customer determines the customer’s chances of winning a prize, a manager Hypothesis that, during sweepstakes time, the proportion of Bank America traveler’s check buyers who buy more than $2,500 worth of checks will be at least 10% higher than the proportion of traveler’s check buyers who buy more than $2,500 worth of checks when there are no sweepstakes. A random sample of 300 traveler’s check buyers, taken when the sweepstakes are on, reveals that 120 of these people bought checks for more than $2,500. A random sample of 700 traveler’s check buyers, taken when no sweepstakes prizes are offered, reveals that 140 of these people checks for more than $2,500. Conduct the Hypothesis test.

Ho: p1- p2 ≤ 0.10

H1: p1- p2 > 0.10

Formula:-

p̂1 - p̂2 ± z /2 √ { p̂1 (1- p̂1) / n1 + p̂2 (1- p̂2) / n2 }

N1 = 300 p̂1 = 120/300 = 0.4

N2 = 700 p̂2 = 140/700 = 0.2

Therefore,

0.4-0.2 ± 1.96 √(0.4)(0.6)/300 + (0.2)(0.8)/700

0.2 ± 1.96 * 0.032

[ 0.137 , 0.263 ]

CONCLUSION----- The manager may be 95% confident that the difference

between the two proportions of interest is anywhere from 0.137 to 0.263.

Q.6 A study was undertaken by Montgomery Securities to assess average labor and

materials costs incurred by Chrysler and General Motors’ (GM) in building a typical four door,

intermediate size car. The reported avg. cost for Chrysler was $9,500 and for GM it was $9,780.

Suppose that these data are based on random samples of 25 cars for each company, and

suppose that both SD are equal to $ 1,500. Test the hypothesis that the avg. GM car of this type

is more expensive to build than the average Chrysler car of the same type. Use α = 0.01.

Give a 99% Confidence interval for the difference between the avg. cost incurred by GM in

building the car and the avg. cost incurred by Chrysler in building the same type of car.

Formula:-

x1- x2 ± z /2

x1= 9,780 x2= 9,500

n1 = 25 n2 = 25

SD1 & SD2 = 1,500

9780- 9500 ± 2.58 √15002/25 + 15002/25

280 ± 2.58 √1,80,000

280 ± 2.58* 424.26

[-814.59, 1374.59]

The Confidence interval shows the non rejection of the null hypothesis. With 99% confidence

we can say that the difference between the avg. cost incurred by GM in building the car and the

avg. cost incurred by Chrysler in building the same type of car is different or higher.

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