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.

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:

$H_0:\text{Median}=220\text{ milligram percent}$

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

Alternative hypothesis:

$H_1:\text{Median}\ne 220\text{ milligram percent}$

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	$\text{Difference}=\text{Serum cholesterol}-220$
210	–10
206	–14
215	–5
223	3
200	–20
250	30
220	0
187	–33
186	–34
193	–27

The sample size is,

$\begin{array}{l}n=\text{ sum of }(+)\text{ and }(-)\text{ signs}\hfill \\ =2+7\hfill \\ =9\hfill \end{array}$

Critical value:

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

Test statistic:

Condition:

If $n\le 25,$ the test statistic for sign test is $x$.

If $n>25$, the test statistic for sign test is $z=\frac{\left(x+0.5\right)-\frac{n}{2}}{\frac{\sqrt{n}}{2}}$

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.

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:

$H_0:\text{Median}=120\text{ millimeters of mercury}$

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

Alternative hypothesis:

$H_1:\text{Median}\ne 120\text{ millimeters of mercury}$

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	$\text{Difference}=\text{Systolic pressure}-120$
129	9
131	11
115	–5
122	2
119	–1
131	11
121	1
117	–3
142	22
123	3

The sample size is,

$\begin{array}{l}n=\text{ sum of }(+)\text{ and }(-)\text{ signs}\hfill \\ =\text{ 7}+3\hfill \\ =10\hfill \end{array}$

Critical value:

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

Test statistic:

Condition:

If $n\le 25,$ the test statistic for sign test is $x$.

If $n>25$, the test statistic for sign test is $z=\frac{\left(x+0.5\right)-\frac{n}{2}}{\frac{\sqrt{n}}{2}}$

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.

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:

$H_0:\text{Median}=100$

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

Alternative hypothesis:

$H_1:\text{Median}\ne 100$

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	$\text{Difference}=IQ-100$
106	6
99	–1
101	1
121	21
99	–1
95	–5
100	0
121	21
103	3
127	27

The sample size is,

$\begin{array}{l}n=\text{ sum of }(+)\text{ and }(-)\text{ signs}\hfill \\ =\text{ 6}+3\hfill \\ =9\hfill \end{array}$

Critical value:

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

Test statistic:

Condition:

If $n\le 25,$ the test statistic for sign test is $x$.

If $n>25$, the test statistic for sign test is $z=\frac{\left(x+0.5\right)-\frac{n}{2}}{\frac{\sqrt{n}}{2}}$

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.

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:

$H_0:\text{Median}=140\text{ mEq/l}$

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

Alternative hypothesis:

$H_1:\text{Median}\ne 140\text{ mEq/l}$

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	$\text{Difference}=\text{Sodium}-140$
136	–4
140	0
144	4
132	–8
139	–1
146	6
143	3
146	6
131	–9
145	5

The sample size is,

$\begin{array}{l}n=\text{ sum of }(+)\text{ and }(-)\text{ signs}\hfill \\ =\text{ 5}+4\hfill \\ =9\hfill \end{array}$

Critical value:

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

Test statistic:

Condition:

If $n\le 25,$ the test statistic for sign test is $x$.

If $n>25$, the test statistic for sign test is $z=\frac{\left(x+0.5\right)-\frac{n}{2}}{\frac{\sqrt{n}}{2}}$

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.