Loose Leaf for Elementary Statistics: A Step By Step Approach
Loose Leaf for Elementary Statistics: A Step By Step Approach
10th Edition
ISBN: 9781260152821
Author: Bluman, Allan G.
Publisher: McGraw-Hill Education
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Chapter 13, Problem 1DA

a.

To determine

To test: The claim that the sample is from a population with a median equal to 220 milligram percent of the serum cholesterol by using the sign test.

a.

Expert Solution
Check Mark

Answer to Problem 1DA

There is no evidence to reject the claim that the sample is from a population with a median equal to 220 milligram percent of the serum cholesterol.

Explanation of Solution

Given info:

The data shows that the different measures of the people.

Calculation:

State the null and alternative hypothesis.

Null hypothesis:

 H0:Median=220 milligram percent

That is, the sample is from a population with a median equal to 220 milligram percent.

Alternative hypothesis:

H1:Median220 milligram percent

That is, the sample is from a population with a median not equal to 220 milligram percent.

Determine sample size n:

Subtract the hypothesized median of 220 from with the sample of serum cholesterol.

Serum cholesterol

Difference=Serum cholesterol220
210 –10
206 –14
215 –5
223 3
200 –20
250 30
220 0
187 –33
186 –34
193 –27

The sample size is,

n= sum of (+) and () signs =2+7    = 9

Critical value:

From Table J” Critical values for Sign test” for n=9 at α=0.05 in two-tailed test, the critical value is 1.

Test statistic:

Condition:

If n25, the test statistic for sign test is x .

If n>25 , the test statistic for sign test is z=(x+0.5)n2n2

Where x is the smaller number of (+) & () and n is the sample size.

Here, the number of (+) sign in the table is 2 and number of () sign in the table is 7 also the samples size is less than 25. That is 9<25 .

Here, test statistic represents the smaller of the number of positive and negative signs. Thus, the test statistic is 2.

Decision:

If test statistic less than or equal to the critical value then reject the null hypothesis.

Conclusion:

The test statistic is 2 and the critical value is 1.

Here, test statistic is greater than the critical value.

That is, 2>1 .

Thus, do not reject the null hypothesis.

Hence, there is no evidence to reject the claim that the sample is from a population with a median equal to 220 milligram percent of the serum cholesterol.

b.

To determine

To test: The claim that the sample is from a population with a median equal to 120 millimeters of mercury of the systolic pressure by using the sign test.

b.

Expert Solution
Check Mark

Answer to Problem 1DA

There is no evidence to reject the claim that the sample is from a population with a median equal to 120 millimeters of mercury of the systolic pressure.

Explanation of Solution

Calculation:

State the null and alternative hypothesis.

Null hypothesis:

 H0:Median=120 millimeters of mercury

That is, the sample is from a population with a median equal to 120 millimeters of mercury.

Alternative hypothesis:

H1:Median120 millimeters of mercury

That is, the sample is from a population with a median not equal to 120 millimeters of mercury.

Determine sample size n:

Subtract the hypothesized median of 120 from with the sample of systolic pressure.

Systolic Pressure

Difference=Systolic pressure120
129 9
131 11
115 –5
122 2
119 –1
131 11
121 1
117 –3
142 22
123 3

The sample size is,

n= sum of (+) and () signs = 7+3    = 10

Critical value:

From Table J” Critical values for Sign test” for n=10 at α=0.05 in two-tailed test, the critical value is 1.

Test statistic:

Condition:

If n25, the test statistic for sign test is x .

If n>25 , the test statistic for sign test is z=(x+0.5)n2n2

Where x is the smaller number of (+) & () and n is the sample size.

Here, the number of (+) sign in the table is 7 and number of () sign in the table is 3 also the samples size is less than 25. That is 10<25 .

Here, test statistic represents the smaller of the number of positive and negative signs. Thus, the test statistic is 3.

Decision:

If test statistic less than or equal to the critical value then reject the null hypothesis.

Conclusion:

The test statistic is 3 and the critical value is 1.

Here, test statistic is greater than the critical value.

That is, 3>1 .

Thus, do not reject the null hypothesis.

Hence, there is no evidence to reject the claim that the sample is from a population with a median equal to 120 millimeters of mercury of the systolic pressure.

c.

To determine

To test: The claim that the sample is from a population with a median equal to 100 for the IQ by using the sign test.

c.

Expert Solution
Check Mark

Answer to Problem 1DA

There is no evidence to reject the claim that the sample is from a population with a median equal to 100 for the IQ.

Explanation of Solution

Calculation:

State the null and alternative hypothesis.

Null hypothesis:

 H0:Median=100

That is, the sample is from a population with a median equal to 100.

Alternative hypothesis:

H1:Median100

That is, the sample is from a population with a median not equal to 100.

Determine sample size n:

Subtract the hypothesized median of 100 from with the sample of IQ.

IQ Difference=IQ100
106 6
99 –1
101 1
121 21
99 –1
95 –5
100 0
121 21
103 3
127 27

The sample size is,

n= sum of (+) and () signs = 6+3    = 9

Critical value:

From Table J” Critical values for Sign test” for n=9 at α=0.05 in two-tailed test, the critical value is 1.

Test statistic:

Condition:

If n25, the test statistic for sign test is x .

If n>25 , the test statistic for sign test is z=(x+0.5)n2n2

Where x is the smaller number of (+) & () and n is the sample size.

Here, the number of (+) sign in the table is 6 and number of () sign in the table is 3 also the samples size is less than 25. That is 9<25 .

Here, test statistic represents the smaller of the number of positive and negative signs. Thus, the test statistic is 3.

Decision:

If test statistic less than or equal to the critical value then reject the null hypothesis.

Conclusion:

The test statistic is 3 and the critical value is 1.

Here, test statistic is greater than the critical value.

That is, 3>1 .

Thus, do not reject the null hypothesis.

Hence, there is no evidence to reject the claim that the sample is from a population with a median equal to 100 for the IQ.

d.

To determine

To test: The claim that the sample is from a population with a median equal to 140 for the sodium by using the sign test.

d.

Expert Solution
Check Mark

Answer to Problem 1DA

There is no evidence to reject the claim that the sample is from a population with a median equal to 140 mEq/l for the sodium level.

Explanation of Solution

Calculation:

State the null and alternative hypothesis.

Null hypothesis:

 H0:Median=140 mEq/l

That is, the sample is from a population with a median equal to 140 mEq/l.

Alternative hypothesis:

H1:Median140 mEq/l

That is, the sample is from a population with a median not equal to 140 mEq/l.

Determine sample size n:

Subtract the hypothesized median of 140 from with the sample of sodium.

Sodium Difference=Sodium140
136 –4
140 0
144 4
132 –8
139 –1
146 6
143 3
146 6
131 –9
145 5

The sample size is,

n= sum of (+) and () signs = 5+4    = 9

Critical value:

From Table J” Critical values for Sign test” for n=9 at α=0.05 in two-tailed test, the critical value is 1.

Test statistic:

Condition:

If n25, the test statistic for sign test is x .

If n>25 , the test statistic for sign test is z=(x+0.5)n2n2

Where x is the smaller number of (+) & () and n is the sample size.

Here, the number of (+) sign in the table is 5 and number of () sign in the table is 4 also the samples size is less than 25. That is 9<25 .

Here, test statistic represents the smaller of the number of positive and negative signs. Thus, the test statistic is 4.

Decision:

If test statistic less than or equal to the critical value then reject the null hypothesis.

Conclusion:

The test statistic is 4 and the critical value is 1.

Here, test statistic is greater than the critical value.

That is, 4>1 .

Thus, do not reject the null hypothesis.

Hence, there is no evidence to reject the claim that the sample is from a population with a median equal to mEq/l for the sodium level.

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Chapter 13 Solutions

Loose Leaf for Elementary Statistics: A Step By Step Approach

Ch. 13.1 - Prob. 10ECh. 13.1 - Prob. 11ECh. 13.1 - Prob. 12ECh. 13.2 - Clean Air An environmentalist suggests that the...Ch. 13.2 - Exercises 132 1. Why is the sign test the simplest...Ch. 13.2 - Prob. 2ECh. 13.2 - Prob. 3ECh. 13.2 - Prob. 4ECh. 13.2 - For Exercises 5 through 20, perform these steps....Ch. 13.2 - Prob. 6ECh. 13.2 - AID: 1825 | 12/01/2018 7. 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