The Practice of Statistics for AP - 4th Edition
The Practice of Statistics for AP - 4th Edition
4th Edition
ISBN: 9781429245593
Author: Starnes, Daren S., Yates, Daniel S., Moore, David S.
Publisher: Macmillan Higher Education
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Chapter 10.1, Problem 28E
Solution

a)

The importance of researchers chose a rural highway so that cars would be separated rather than in clusters.

Given:A significance test is carried out to test the claim that the proportion of speeding drivers differ significantly when radar us being used and when it isn’t.

Explanation:The researchers chose a rural highway so that cars would be separated rather than in clusters, because some cars might slow down when they see other cars slowing. If a driver change the speed because of another car’s speed, then the observations are no longer considered as independent. We can use two sample Z test for two proportions only when the observations are independent.

b)

To test the claim that the proportion of speeding drivers differ significantly when radar is being used and when it isn’t.

Given:

The number of vehicles with no radar and over 65 mph x1=5690

The number of vehicles with radar and over 65 mph x2=1051

The number of vehicles with no radar n1=12931

The number of vehicles with radar n2=3285

The Minitab output is also given

P1 be the proportion vehicles with no radar and over 65 mph is 0.440028

P2 be the proportion vehicles with radar and over 65 mph is 0.319939

The estimate for difference : 0.120089

95% confidence interval for difference : (0.101988, 0.138189)

The null and alternate hypothesis are H0:p1=p2Ha:p1p2

Z value is 12.47 and p value is 0.000

Concept used: The conditions to satisfy to run two sample Z test for difference in proportions are

1) The sample should be random

2) np = 10

3) n(1-p) = 10

4) The observations in each group should be independent.

Calculation:

  n1p1=12931*0.440028=5690n1p1>10n1(1p1)=12931*(10.440028)=7241n1(1p1)>10n2p2=3285*0.319939=1051n2p2>10n2(1p2)=3285*(10.319939)=2234n2(1p2)>10

Both samples are selected randomly.

As there would be at least 32850 cars when there is radar and 129310 cars when there is no radar, we consider both samples are less than 10% of respective populations. So individual observations in each group are also considered to be independent.

All conditions are met so we can use two sample Z test for difference in proportions.

From Minitab output the test statistic is Z=12.47 and pvalue=0.000

Since the p value is less than the level of significance we have sufficient evidence at 5% level of significance to reject the null hypothesis.

Conclusion:

At 5% level of significance we have strong evidence to conclude that there is a difference in proportion of speeders between when there is radar present and when it is not.

Chapter 10 Solutions

The Practice of Statistics for AP - 4th Edition

Ch. 10.1 - Prob. 4ECh. 10.1 - Prob. 5ECh. 10.1 - Prob. 6ECh. 10.1 - Prob. 7ECh. 10.1 - Prob. 8ECh. 10.1 - Prob. 9ECh. 10.1 - Prob. 10ECh. 10.1 - Prob. 11ECh. 10.1 - Prob. 12ECh. 10.1 - Prob. 13ECh. 10.1 - Prob. 14ECh. 10.1 - Prob. 15ECh. 10.1 - Prob. 16ECh. 10.1 - Prob. 17ECh. 10.1 - Prob. 18ECh. 10.1 - Prob. 19ECh. 10.1 - Prob. 20ECh. 10.1 - Prob. 21ECh. 10.1 - Prob. 22ECh. 10.1 - Prob. 23ECh. 10.1 - Prob. 24ECh. 10.1 - Prob. 25ECh. 10.1 - Prob. 26ECh. 10.1 - Prob. 27ECh. 10.1 - Prob. 28ECh. 10.1 - Prob. 29ECh. 10.1 - Prob. 30ECh. 10.1 - Prob. 31ECh. 10.1 - Prob. 32ECh. 10.1 - Prob. 33ECh. 10.1 - Prob. 34ECh. 10.2 - Prob. 1.1CYUCh. 10.2 - Prob. 1.2CYUCh. 10.2 - Prob. 1.3CYUCh. 10.2 - Prob. 1.4CYUCh. 10.2 - Prob. 2.1CYUCh. 10.2 - Prob. 3.1CYUCh. 10.2 - Prob. 35ECh. 10.2 - Prob. 36ECh. 10.2 - Prob. 37ECh. 10.2 - Prob. 38ECh. 10.2 - Prob. 39ECh. 10.2 - Prob. 40ECh. 10.2 - Prob. 41ECh. 10.2 - Prob. 42ECh. 10.2 - Prob. 43ECh. 10.2 - Prob. 44ECh. 10.2 - Prob. 45ECh. 10.2 - Prob. 46ECh. 10.2 - Prob. 47ECh. 10.2 - Prob. 48ECh. 10.2 - Prob. 49ECh. 10.2 - Prob. 50ECh. 10.2 - Prob. 51ECh. 10.2 - Prob. 52ECh. 10.2 - Prob. 53ECh. 10.2 - Prob. 54ECh. 10.2 - Prob. 55ECh. 10.2 - Prob. 56ECh. 10.2 - Prob. 57ECh. 10.2 - Prob. 58ECh. 10.2 - Prob. 59ECh. 10.2 - Prob. 60ECh. 10.2 - Prob. 61ECh. 10.2 - Prob. 62ECh. 10.2 - Prob. 63ECh. 10.2 - Prob. 64ECh. 10.2 - Prob. 65ECh. 10.2 - Prob. 66ECh. 10.2 - Prob. 67ECh. 10.2 - Prob. 68ECh. 10.2 - Prob. 69ECh. 10.2 - Prob. 70ECh. 10.2 - Prob. 71ECh. 10.2 - Prob. 72ECh. 10.2 - Prob. 73ECh. 10.2 - Prob. 74ECh. 10.2 - Prob. 75ECh. 10.2 - Prob. 76ECh. 10 - Prob. 1CRECh. 10 - Prob. 2CRECh. 10 - Prob. 3CRECh. 10 - Prob. 4CRECh. 10 - Prob. 5CRECh. 10 - Prob. 6CRECh. 10 - Prob. 7CRECh. 10 - Prob. 8CRECh. 10 - Prob. 9CRECh. 10 - Prob. 10CRECh. 10 - Prob. 1PTCh. 10 - Prob. 2PTCh. 10 - Prob. 3PTCh. 10 - Prob. 4PTCh. 10 - Prob. 5PTCh. 10 - Prob. 6PTCh. 10 - Prob. 7PTCh. 10 - Prob. 8PTCh. 10 - Prob. 9PTCh. 10 - Prob. 10PTCh. 10 - Prob. 11PTCh. 10 - Prob. 12PTCh. 10 - Prob. 13PTCh. 10 - Prob. 1PT3Ch. 10 - Prob. 2PT3Ch. 10 - Prob. 3PT3Ch. 10 - Prob. 4PT3Ch. 10 - Prob. 5PT3Ch. 10 - Prob. 6PT3Ch. 10 - Prob. 7PT3Ch. 10 - Prob. 8PT3Ch. 10 - Prob. 9PT3Ch. 10 - Prob. 10PT3Ch. 10 - Prob. 11PT3Ch. 10 - Prob. 12PT3Ch. 10 - Prob. 13PT3Ch. 10 - Prob. 14PT3Ch. 10 - Prob. 15PT3Ch. 10 - Prob. 16PT3Ch. 10 - Prob. 17PT3Ch. 10 - Prob. 18PT3Ch. 10 - Prob. 19PT3Ch. 10 - Prob. 20PT3Ch. 10 - Prob. 21PT3Ch. 10 - Prob. 22PT3Ch. 10 - Prob. 23PT3Ch. 10 - Prob. 24PT3Ch. 10 - Prob. 25PT3Ch. 10 - Prob. 26PT3Ch. 10 - Prob. 27PT3Ch. 10 - Prob. 28PT3Ch. 10 - Prob. 29PT3Ch. 10 - Prob. 30PT3Ch. 10 - Prob. 31PT3Ch. 10 - Prob. 32PT3Ch. 10 - Prob. 33PT3Ch. 10 - Prob. 34PT3Ch. 10 - Prob. 35PT3
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