Question

The number of cell phones per 100 residents in countries in Europe is given in Table #1 for the year 2010. The number of cell phones per 100 residents in countries of the Americas is given in Table #2 also for the year 2010 ("Population reference bureau," 2013).

Table #1: Number of Cell Phones per 100 Residents in Europe

100	76	100	130	75	84
112	84	138	133	118	134
126	188	129	93	64	128
124	122	109	121	127	152
96	63	99	95	151	147
123	95	67	67	118	125
110	115	140	115	141	77
98	102	102	112	118	118
54	23	121	126	47

Table #2: Number of Cell Phones per 100 Residents in the Americas

158	117	106	159	53	50
78	66	88	92	42	3
150	72	86	113	50	58
70	109	37	32	85	101
75	69	55	115	95	73
86	157	100	119	81	113
87	105	96

Let ?₁= mean number of cell phones per 100 residents in countries of Europe. Let ?₂ = mean number of cell phones per 100 residents in countries of the Americas. Find the 98% confidence interval for the difference in mean number of cell phones per 100 residents in Europe and the Americas.

(i) For the sample from population with mean = ?1 : Determine sample mean x¯1 and sample standard deviation s1i  For the sample from population with mean = ?1 :

Enter sample mean in decimal form to nearest ten-thousandth, then comma, then sample standard deviation in decimal form to nearest ten-thousandth. Examples of correctly entered answers:

13.2027,2.3471

0.2700,0.0006

-10.3150,0.0790

   (ii) For the sample from the population with mean = ?2 : Determine sample mean x¯2 and sample standard deviation s2ii  For the sample from the population with mean = ?2 :      Determine sample mean x¯2 and sample standard deviation s2 {"version":"1.1","math":"<math xmlns="http://www.w3.org/1998/Math/MathML"><mstyle mathsize="14px"><mrow><mfenced><mi mathvariant="bold">ii</mi></mfenced><mo mathvariant="bold">&#xA0;</mo><mo mathvariant="bold">&#xA0;</mo><mi mathvariant="bold">For</mi><mo mathvariant="bold">&#xA0;</mo><mi mathvariant="bold">the</mi><mo mathvariant="bold">&#xA0;</mo><mi mathvariant="bold">sample</mi><mo mathvariant="bold">&#xA0;</mo><mi mathvariant="bold">from</mi><mo mathvariant="bold">&#xA0;</mo><mi mathvariant="bold">the</mi><mo mathvariant="bold">&#xA0;</mo><mi mathvariant="bold">population</mi><mo mathvariant="bold">&#xA0;</mo><mi mathvariant="bold">with</mi><mo mathvariant="bold">&#xA0;</mo><mi mathvariant="bold">mean</mi><mo mathvariant="bold">&#xA0;</mo><mo mathvariant="bold">=</mo><mo mathvariant="bold">&#xA0;</mo><msub><mi mathvariant="bold-italic">&#x3BC;</mi><mn mathvariant="bold">2</mn></msub><mo mathvariant="bold">&#xA0;</mo><mo mathvariant="bold">:</mo><mspace linebreak="newline"></mspace><mo mathvariant="bold">&#xA0;</mo><mo mathvariant="bold">&#xA0;</mo><mo mathvariant="bold">&#xA0;</mo><mo mathvariant="bold">&#xA0;</mo><mo mathvariant="bold">&#xA0;</mo><mo mathvariant="bold">&#xA0;</mo><mi mathvariant="bold">Determine</mi><mo mathvariant="bold">&#xA0;</mo><mi mathvariant="bold">sample</mi><mo mathvariant="bold">&#xA0;</mo><mi mathvariant="bold">mean</mi><mo mathvariant="bold">&#xA0;</mo><msub><mover><mi mathvariant="bold">x</mi><mo mathvariant="bold">&#xAF;</mo></mover><mn mathvariant="bold">2</mn></msub><mo mathvariant="bold">&#xA0;</mo><mi mathvariant="bold">and</mi><mo mathvariant="bold">&#xA0;</mo><mi mathvariant="bold">sample</mi><mo mathvariant="bold">&#xA0;</mo><mi mathvariant="bold">standard</mi><mo mathvariant="bold">&#xA0;</mo><mi mathvariant="bold">deviation</mi><mo mathvariant="bold">&#xA0;</mo><msub><mi mathvariant="bold-italic">s</mi><mn mathvariant="bold">2</mn></msub></mrow></mstyle></math>"}

Enter sample mean in decimal form to nearest ten-thousandth, then comma, then sample standard deviation in decimal form to nearest ten-thousandth. Examples of correctly entered answers:

13.2027,2.3471

0.2700,0.0006

-10.3150,0.0790

(iii) Enter the level of significance ? used for this test:

Enter in decimal form. Examples of correctly entered answers: 0.01    0.02    0.05    0.10

(iv) Determine degrees of freedom df :

Enter value rounded DOWN to nearest integer.

(v) Determine t - score associated with critical value: t_c

Enter in decimal form to nearest thousandth. Examples of correctly entered answers:

0.0011    0.020     0.500     0.371 2.000

(vi) Determine "error bound of the mean" E

Enter value in decimal form rounded to nearest thousandth. Examples of correctly entered answers:

2.014      3.070        16.000        110.019

(vii) Determine confidence interval estimate of the difference ?₁ – ?₂:

Enter lower bound value to nearest ten-thousandth, followed by < , followed by "?1-?2" for difference, followed by <, followed by upper bound value to nearest ten-thousandth. No spaces between any characters. Do not use italics, but if mathematically necessary use negative signs. Examples of correctly entered answers:

0.0077<?1-?2<0.0780

13.1012<?1-?2<18.3941

-9.7300<?1-?2<-3.0008

(viii) Using the confidence interval, select the most correct description of the result of the survey:

A. We estimate with 98% confidence that the sample mean number of cell phones per 100 residents in countries of Europe is anywhere from 4.3641 to 37.5276 more than the mean number of cell phones per 100 residents in countries of the Americas.

B. We estimate with 98% confidence that the true mean number of cell phones per 100 residents in countries of Europe is anywhere from 4.3641 to 37.5276 more than the mean number of cell phones per 100 residents in countries of the Americas.

C. We estimate with 98% confidence that the true proportion of cell phones per 100 residents in countries of Europe is anywhere from 4.3641 to 37.5276 more than the mean number of cell phones per 100 residents in countries of the Americas.

D. We estimate with 98% confidence that the mean number of cell phones per 100 residents in countries of Europe is anywhere from 4.3641 to 37.5276 less than the mean number of cell phones per 100 residents in countries of the Americas.

Answer 1

(i) For the sample from population with mean = ?1 : Determine sample mean x¯1 and samplestandard deviation s1

mean of sample 1 = 108.1509434

standard deviation of sample 1 = 29.96497714

So, the answer is,

108.1509,29.9650

(ii) For the sample from the population with mean = ?2 : Determine sample mean x¯2 andsample standard deviation s2

mean of sample 2 = 87.20512821

standard deviation of sample 1 = 35.15542893

So, the answer is,

87.2051,35.1554

(iii) Enter the level of significance ? used for this test:

For 98% confidence interval,

1 - 2 = 0.98

=> = (1 - 0.98)/2 = 0.01

Level of significance is 0.01

(iv) Determine degrees of freedom df :

Let s1 and s2 be the sample standard deviations for sample 1 and sample 2 respectively.

And let n1 and n2 be the sample size for sample 1 and sample 2 respectively.

DF = (s1²/n1 + s2²/n2)² / { [ (s1² / n1)² / (n1 - 1) ] + [ (s2² / n2)² / (n2 - 1) ] }

= (29.9650²/53 + 35.1554²/39)² / { [ (29.9650² / 53)² / (53 - 1) ] + [ (35.1554² / 39)² / (39 - 1) ] }

= 74 (Rounding to nearest integer)

(v) Determine t - score associated with critical value: tc

Standard error (SE) of the sampling distribution.

SE = sqrt[ (s1²/n1) + (s2²/n2) ]

= sqrt[ (29.9650²/53) + (35.1554²/39) ]

= 6.973616798

t statistic = (Difference in means) / SE

= (108.1509 - 87.2051) / 6.973616798

= 3.003

(vi) Determine "error bound of the mean" E

t for 98% confidence inetrval (level of significance 0.01) with DF = 74 is 2.3778

Margin of error = t * SE

= 2.3778 * 6.973616798

= 16.582

(vii) Determine confidence interval estimate of the difference ?1 – ?2:

(Difference in means) - Margin of error <?1-?2< (Difference in means) + Margin of error

(108.1509 - 87.2051) - 16.582 <?1-?2< (108.1509 - 87.2051) + 16.582

4.3638 <?1-?2< 37.5278

(viii) Using the confidence interval, select the most correct description of the result of the survey:

B. We estimate with 98% confidence that the true mean number of cell phones per 100 residents in countries of Europe is anywhere from 4.3641 to 37.5276 more than the mean number of cell phones per 100 residents in countries of the Americas.