PRACTICE OF STATISTICS F/AP EXAM
PRACTICE OF STATISTICS F/AP EXAM
6th Edition
ISBN: 9781319113339
Author: Starnes
Publisher: MAC HIGHER
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Chapter 10.1, Problem 29E

(a)

To determine

To state appropriate hypotheses for performing a significance test and also define the parameters.

(a)

Expert Solution
Check Mark

Answer to Problem 29E

The hypotheses is:

  H0:p1=p2Ha:p1p2

Explanation of Solution

It is given that the researcher wants to study about texting and driving by asking different versions of questions.

So, the given claim is that: the proportion are different and proportion is affected by additional information in the question.

Now, we have to find out the appropriate hypotheses for performing a significance test.

Thus, the claim is either the null hypothesis or the alternative hypothesis. The null hypothesis states that the population proportions are equal. If the null hypothesis is the claim then the alternative hypothesis states the opposite of the null hypothesis.

Therefore, the appropriate hypotheses for this is:

  H0:p1=p2Ha:p1p2

Where we have,

  p1= the proportion of shoppers who admit to texting while driving when hearing version A.

  p2= the proportion of shoppers who admit to texting while driving when hearing version B.

(b)

To determine

To explain why you should not use the method of this section to calculate the P -value.

(b)

Expert Solution
Check Mark

Explanation of Solution

Given:

It is given that the researcher wants to study about texting and driving by asking different versions of questions.

  x1=16x2=12n1=25n2=25

Conditions to be satisfied:

There are three conditions to be satisfied:

Random: It is satisfied because the shoppers are independent random samples.

Independent: It is satisfied because the 25 shoppers are less than 10% of all shoppers

Normal: It is not satisfied because there are 16 successes and 2516=9 failures in the first sample, which is not at least ten.

Thus, all the conditions are not satisfied since the large costs is not satisfied and it is not appropriate to use hypothesis test for testing a claim. Thus, you should not use the method of this section to calculate the P -value.

(c)

To determine

To find out what is the estimated P -value.

(c)

Expert Solution
Check Mark

Answer to Problem 29E

The estimated P values is 0.35 .

Explanation of Solution

Given:

It is given that the researcher wants to study about texting and driving by asking different versions of questions.

  x1=16x2=12n1=25n2=25

Hypotheses:

So, the given claim is that: the proportion are different and proportion is affected by additional information in the question.

Now, we have to find out the appropriate hypotheses for performing a significance test.

Thus, the claim is either the null hypothesis or the alternative hypothesis. The null hypothesis states that the population proportions are equal. If the null hypothesis is the claim then the alternative hypothesis states the opposite of the null hypothesis.

Therefore, the appropriate hypotheses for this is:

  H0:p1=p2Ha:p1p2

Where we have,

  p1= the proportion of shoppers who admit to texting while driving when hearing version A.

  p2= the proportion of shoppers who admit to texting while driving when hearing version B.

Calculation:

The sample proportion is the number of successes divided by the sample size. Then, we have,

  p^1=x1n1=1625=0.64p^2=x2n2=1225=0.48

Now, we will calculate the difference of the sample proportions:

  p^1p^2=0.640.48=0.16

The P-value is the probability of obtaining the value of the test statistics or a value more extreme assuming that the null hypothesis is true. Thus, we have,

There are ten dots at 0.16 and four dots to the right of 0.16 , which implies that 10+4=14 of the 100 dots in the dot plot corresponds to the difference of sample proportions of at least 0.16 .

Also, there are 13 dots at 0.16 and eight dots to the left of 0.16 , which implies that 13+8=21 of the 100 dots in the dot plot corresponds with a difference of sample proportions of at most 0.16 .

Thus, 14+21=35 of the 100 dots in the dot plot corresponds with a difference of sample proportions that is at least 0.16 from the difference of zero when the sample proportions are the same.

  P-value=35100=0.35

Thus, the estimated P values is 0.35 .

(d)

To determine

To explain what conclusion what you draw in this context.

(d)

Expert Solution
Check Mark

Explanation of Solution

From part (c) we have that,

The appropriate hypotheses for this is:

  H0:p1=p2Ha:p1p2

Where we have,

  p1= the proportion of shoppers who admit to texting while driving when hearing version A.

  p2= the proportion of shoppers who admit to texting while driving when hearing version B.

Calculation:

The sample proportion is the number of successes divided by the sample size. Then, we have,

  p^1=x1n1=1625=0.64p^2=x2n2=1225=0.48

Now, we will calculate the difference of the sample proportions:

  p^1p^2=0.640.48=0.16

The P-value is the probability of obtaining the value of the test statistics or a value more extreme assuming that the null hypothesis is true. Thus, we have,

There are ten dots at 0.16 and four dots to the right of 0.16 , which implies that 10+4=14 of the 100 dots in the dot plot corresponds to the difference of sample proportions of at least 0.16 .

Also, there are 13 dots at 0.16 and eight dots to the left of 0.16 , which implies that 13+8=21 of the 100 dots in the dot plot corresponds with a difference of sample proportions of at most 0.16 .

Thus, 14+21=35 of the 100 dots in the dot plot corresponds with a difference of sample proportions that is at least 0.16 from the difference of zero when the sample proportions are the same.

  P-value=35100=0.35

Thus, if the P-value is smaller than the significance level, then we will reject the null hypothesis, thus, we have,

  P>0.05Fail to Reject H0

Thus, we conclude that there is no convincing evidence that the proportion of shoppers who admitted to texting while driving is different for the two versions of the question asked.

Chapter 10 Solutions

PRACTICE OF STATISTICS F/AP EXAM

Ch. 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. 10.1 - Prob. 35ECh. 10.1 - 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. 10.3 - Prob. 75ECh. 10.3 - Prob. 76ECh. 10.3 - Prob. 77ECh. 10.3 - Prob. 78ECh. 10.3 - Prob. 79ECh. 10.3 - Prob. 80ECh. 10.3 - Prob. 81ECh. 10.3 - Prob. 82ECh. 10.3 - Prob. 83ECh. 10.3 - Prob. 84ECh. 10.3 - Prob. 85ECh. 10.3 - Prob. 86ECh. 10.3 - Prob. 87ECh. 10.3 - Prob. 88ECh. 10.3 - Prob. 89ECh. 10.3 - Prob. 90ECh. 10.3 - Prob. 91ECh. 10.3 - Prob. 92ECh. 10.3 - Prob. 93ECh. 10.3 - Prob. 94ECh. 10.3 - Prob. 95ECh. 10.3 - Prob. 96ECh. 10.3 - Prob. 97ECh. 10.3 - Prob. 98ECh. 10.3 - Prob. 99ECh. 10.3 - Prob. 100ECh. 10.3 - Prob. 101ECh. 10.3 - Prob. 102ECh. 10 - Prob. R10.1RECh. 10 - Prob. R10.2RECh. 10 - Prob. R10.3RECh. 10 - Prob. R10.4RECh. 10 - Prob. R10.5RECh. 10 - Prob. R10.6RECh. 10 - Prob. R10.7RECh. 10 - Prob. T10.1SPTCh. 10 - Prob. T10.2SPTCh. 10 - Prob. T10.3SPTCh. 10 - Prob. T10.4SPTCh. 10 - Prob. T10.5SPTCh. 10 - Prob. T10.6SPTCh. 10 - Prob. T10.7SPTCh. 10 - Prob. T10.8SPTCh. 10 - Prob. T10.9SPTCh. 10 - Prob. T10.10SPTCh. 10 - Prob. T10.11SPTCh. 10 - Prob. T10.12SPTCh. 10 - Prob. T10.13SPTCh. 10 - Prob. AP3.1CPTCh. 10 - Prob. AP3.2CPTCh. 10 - Prob. AP3.3CPTCh. 10 - Prob. AP3.4CPTCh. 10 - Prob. AP3.5CPTCh. 10 - Prob. AP3.6CPTCh. 10 - Prob. AP3.7CPTCh. 10 - Prob. AP3.8CPTCh. 10 - Prob. AP3.9CPTCh. 10 - Prob. AP3.10CPTCh. 10 - Prob. AP3.11CPTCh. 10 - Prob. AP3.12CPTCh. 10 - Prob. AP3.13CPTCh. 10 - Prob. AP3.14CPTCh. 10 - Prob. AP3.15CPTCh. 10 - Prob. AP3.16CPTCh. 10 - Prob. AP3.17CPTCh. 10 - Prob. AP3.18CPTCh. 10 - Prob. AP3.19CPTCh. 10 - Prob. AP3.20CPTCh. 10 - Prob. AP3.21CPTCh. 10 - Prob. AP3.22CPTCh. 10 - Prob. AP3.23CPTCh. 10 - Prob. AP3.24CPTCh. 10 - Prob. AP3.25CPTCh. 10 - Prob. AP3.26CPTCh. 10 - Prob. AP3.27CPTCh. 10 - Prob. AP3.28CPTCh. 10 - Prob. AP3.29CPTCh. 10 - Prob. AP3.30CPTCh. 10 - Prob. AP3.31CPTCh. 10 - Prob. AP3.32CPTCh. 10 - Prob. AP3.33CPTCh. 10 - Prob. AP3.34CPTCh. 10 - Prob. AP3.35CPT
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