Beginning Statistics, 2nd Edition
Beginning Statistics, 2nd Edition
2nd Edition
ISBN: 9781932628678
Author: Carolyn Warren; Kimberly Denley; Emily Atchley
Publisher: Hawkes Learning Systems
Question
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Chapter 11.3, Problem 9E
To determine

(a)

Null hypothesis and alternative hypothesis.

Expert Solution
Check Mark

Answer to Problem 9E

Solution:

The required hypothesis is,

H0: μd10

H1: μd>10

Explanation of Solution

Given:

A local school district is looking at adopting a new textbook that, according to the publishers, will increase standardized test scores of second graders by more than 10 points, on average. Never willing to believe a publisher’s claim without evidence to support it, the school board decides to test the claim. The school board chooses two second-grade classes for the study. One class was assigned the new textbook and the other class used the traditional textbook. Eight children from each class were then paired based on demographics and the ability levels. The following table lists the standardized test scores for the pairs.

New book 78 82 90 67 79 83 89 93
Old book 67 70 79 54 68 71 78 82

Calculation:

Let the test scores of second graders studying old book be population 1 and the test scores of second graders studying new book be population 2. Let μ1 be the mean test score of population 1 and μ2 mean test score of population 2. And μd=μ2μ1 The publisher claims that the standardized test scores will increase by more than 10 points after adopting the new book. Hence the alternative hypothesis is μd>10 and since the null hypothesis is opposite of the alternative hypothesis then,

Thus, the null hypothesis is stated as,

H0: μd10

Alternative hypothesis

H1: μd>10

To determine

(b)

The type of distribution to use for the test statistics and state the level of significance.

Expert Solution
Check Mark

Answer to Problem 9E

Solution:

For the mean of the paired differences t- distribution is used and the level of significance is α=0.05.

Explanation of Solution

Given:

A local school district is looking at adopting a new textbook that, according to the publishers, will increase standardized test scores of second graders by more than 10 points, on average. Never willing to believe a publisher’s claim without evidence to support it, the school board decides to test the claim. The school board chooses two second-grade classes for the study. One class was assigned the new textbook and the other class used the traditional textbook. Eight children from each class were then paired based on demographics and the ability levels. The following table lists the standardized test scores for the pairs.

New book 78 82 90 67 79 83 89 93
Old book 67 70 79 54 68 71 78 82

Calculation:

When the samples given are dependent over each other, the standard deviation of both the populations are unknown and all the given sample have equal probability of getting selected then t-distribution is used.

Since all the criteria described in the concept above are met, hence t-distribution will be used in this given question.

To determine

(c)

To Calculate:

The sample statistic and test statistic value.

Expert Solution
Check Mark

Answer to Problem 9E

Solution:

The sample statistics are d¯=11.5 and sd=0.7559 . And the test statistic is 5.612.

Explanation of Solution

Given:

A local school district is looking at adopting a new textbook that, according to the publishers, will increase standardized test scores of second graders by more than 10 points, on average. Never willing to believe a publisher’s claim without evidence to support it, the school board decides to test the claim. The school board chooses two second-grade classes for the study. One class was assigned the new textbook and the other class used the traditional textbook. Eight children from each class were then paired based on demographics and the ability levels. The following table lists the standardized test scores for the pairs.

New book 78 82 90 67 79 83 89 93
Old book 67 70 79 54 68 71 78 82

Formula used:

When the standard deviation of the populations are unknown and the samples taken are dependent over each other then the test statistic for the hypothesis test for the mean of the paired differences is given by,

t=d¯μd(sdn)

Where d¯ is mean of the paired differences calculated as,

d¯=din

μd is the value of the mean of the paired differences for the two given populations obtained from null hypothesis.

sd is the standard deviation of the mean of paired differences calculated as,

sd=(did¯)n1

n is the total number of data given,

And t is the test statistic value.

Calculation:

From the given information:

The null hypothesis is,

H0: μd10

The alternative hypothesis is,

H1: μd>10

The level of significance α given is 0.05.

Since hypothesis test is for the mean of the paired differences so the distribution which will be used is t- distribution.

To test for the null hypothesis there is a need to calculate the test statistic value.

To calculate test statistic value, t, sample statistic are required.

Old Book New Book d (did¯) (did¯)2
67 78 11 0.5 0.25
70 82 12 0.5 0.25
79 90 11 0.5 0.25
54 67 13 1.5 2.25
68 79 11 0.5 0.25
71 83 12 0.5 0.25
78 89 11 0.5 0.25
82 93 11 0.5 0.25
Sum 92 4

The paired difference is calculated as,

di=x2x1

Substitute 67 for x1 and 78 for x2 in the above equation.

d1=7867=11

Proceed in the same manner to calculate di for the rest of the data and refer table for the rest of the di values calculated. The size of each sample is 8 that is n=8. Then the mean of the paired differences is calculated as,

d¯=din=(0.5)+0.5+....+(0.5)+(0.5)8=928=11.5

Substitute 11 for d1 and 11.5 for d¯ in (d1d¯).

(d1d¯)=(1111.5)(d1d¯)=0.5

Square both sides of the equation.

(d1d¯)2=(0.5)2(d1d¯)2=0.25

Proceed in the same manner to calculate (did¯)2 for all the 1in for the rest data and refer table for the rest of the (did¯)2 values calculated. Then the value of (did¯)2 is calculated as,

(did¯)2=0.25+0.25+...+0.25+0.25=4

The standard deviation is calculated as,

sd=(did¯)n1

Substitute 4 for (did¯)2 and 9 for n in the above equation

sd=47=0.5714=0.7559

Then the test statistic value is,

t=d¯μd(sdn)

Substitute for 11.5 in d¯, 10 in μd, 0.7559 in sd and 8 in n in the formula mentioned above,

t=11.510(0.75598)=5.612

Thus, the t-statistic value is 5.612.

To determine

(d)

To Explain:

The conclusion and interpret the decision.

Expert Solution
Check Mark

Answer to Problem 9E

Solution:

The null hypothesis is rejected and hence publisher’s claim is right.

Explanation of Solution

Given:

A local school district is looking at adopting a new textbook that, according to the publishers, will increase standardized test scores of second graders by more than 10 points, on average. Never willing to believe a publisher’s claim without evidence to support it, the school board decides to test the claim. The school board chooses two second-grade classes for the study. One class was assigned the new textbook and the other class used the traditional textbook. Eight children from each class were then paired based on demographics and the ability levels. The following table lists the standardized test scores for the pairs.

New book 78 82 90 67 79 83 89 93
Old book 67 70 79 54 68 71 78 82

Calculation:

The null hypothesis, H0 is rejected if:

One-Tail test

ttα for a left tailed test.

ttα for a right tailed test.

Two Tail test

|t|tα/2 for two tailed test.

The degree of freedom is,

df=n1=81=7

The critical value is,

tα=t0.05=1.895

So by looking at the test statistic value and the t-distribution value, the test statistic value is greater than the t value, which by right tailed test implies that H0 is rejected. Hence, there is sufficient evidence to reject H0. And the hence the publisher claim that the test scores will increase by 10 points is right.

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Chapter 11 Solutions

Beginning Statistics, 2nd Edition

Ch. 11.1 - Prob. 11ECh. 11.1 - Prob. 12ECh. 11.1 - Prob. 13ECh. 11.1 - Prob. 14ECh. 11.1 - Prob. 15ECh. 11.1 - Prob. 16ECh. 11.1 - Prob. 17ECh. 11.1 - Prob. 18ECh. 11.1 - Prob. 19ECh. 11.1 - Prob. 20ECh. 11.1 - Prob. 21ECh. 11.2 - Prob. 1ECh. 11.2 - Prob. 2ECh. 11.2 - Prob. 3ECh. 11.2 - Prob. 4ECh. 11.2 - Prob. 5ECh. 11.2 - Prob. 6ECh. 11.2 - Prob. 7ECh. 11.2 - Prob. 8ECh. 11.2 - Prob. 9ECh. 11.2 - Prob. 10ECh. 11.2 - Prob. 11ECh. 11.2 - Prob. 12ECh. 11.2 - Prob. 13ECh. 11.2 - Prob. 14ECh. 11.2 - Prob. 15ECh. 11.2 - Prob. 16ECh. 11.2 - Prob. 17ECh. 11.3 - Prob. 1ECh. 11.3 - Prob. 2ECh. 11.3 - Prob. 3ECh. 11.3 - Prob. 4ECh. 11.3 - Prob. 5ECh. 11.3 - Prob. 6ECh. 11.3 - Prob. 7ECh. 11.3 - Prob. 8ECh. 11.3 - Prob. 9ECh. 11.3 - Prob. 10ECh. 11.4 - Prob. 1ECh. 11.4 - Prob. 2ECh. 11.4 - Prob. 3ECh. 11.4 - Prob. 4ECh. 11.4 - Prob. 5ECh. 11.4 - Prob. 6ECh. 11.4 - Prob. 7ECh. 11.4 - Prob. 8ECh. 11.4 - Prob. 9ECh. 11.4 - Prob. 10ECh. 11.4 - Prob. 11ECh. 11.5 - Prob. 1ECh. 11.5 - Prob. 2ECh. 11.5 - Prob. 3ECh. 11.5 - Prob. 4ECh. 11.5 - Prob. 5ECh. 11.5 - Prob. 6ECh. 11.5 - Prob. 7ECh. 11.5 - Prob. 8ECh. 11.5 - Prob. 9ECh. 11.5 - Prob. 10ECh. 11.5 - Prob. 11ECh. 11.5 - Prob. 12ECh. 11.5 - Prob. 13ECh. 11.5 - Prob. 14ECh. 11.5 - Prob. 15ECh. 11.5 - Prob. 16ECh. 11.5 - Prob. 17ECh. 11.5 - Prob. 18ECh. 11.5 - Prob. 19ECh. 11.5 - Prob. 20ECh. 11.5 - Prob. 21ECh. 11.5 - Prob. 22ECh. 11.5 - Prob. 23ECh. 11.5 - Prob. 24ECh. 11.5 - Prob. 25ECh. 11.5 - Prob. 26ECh. 11.6 - Prob. 1ECh. 11.6 - Prob. 2ECh. 11.6 - Prob. 3ECh. 11.6 - Prob. 4ECh. 11.6 - Prob. 5ECh. 11.6 - Prob. 6ECh. 11.6 - Prob. 7ECh. 11.6 - Prob. 8ECh. 11.6 - Prob. 9ECh. 11.6 - Prob. 10ECh. 11.6 - Prob. 11ECh. 11.6 - Prob. 12ECh. 11.6 - Prob. 13ECh. 11.6 - Prob. 14ECh. 11.6 - Prob. 15ECh. 11.6 - Prob. 16ECh. 11.CR - Prob. 1CRCh. 11.CR - Prob. 2CRCh. 11.CR - Prob. 3CRCh. 11.CR - Prob. 4CRCh. 11.CR - Prob. 5CRCh. 11.CR - Prob. 6CRCh. 11.CR - Prob. 7CRCh. 11.CR - Prob. 8CRCh. 11.CR - Prob. 9CRCh. 11.CR - Prob. 10CR
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