Chapter 7, Problem 7.1P

a. For each situation, find Z(critical).

Alpha ( $\text{α}$)	Form of the Test	Z(critical)
0.05	One-tailed
0.10	Two-tailed
0.06	Two-tailed
0.01	One-tailed
0.02	Two-tailed

b. For each situation, find the critical t score.

Alpha ( $\text{α}$)	Form of the Test	N	t(critical)
0.10	Two-tailed	31
0.02	Two-tailed	24
0.01	Two-tailed	121
0.01	One-tailed	31
0.05	One-tailed	61

c. Compute the appropriate test statistic (Z or t) for each situations.

	Population	Sample	Z(obtained) or t(obtained)
1.	$\begin{array}{l}\mu =2.40\\ \sigma =0.75\end{array}$	$\begin{array}{l}\bar{X}=0.20\\ N=200\end{array}$
2.	$\mu =17.1$	$\begin{array}{l}\bar{X}=16.8\\ s=0.9\\ N=45\end{array}$
3.	$\mu =10.2$	$\begin{array}{l}\bar{X}=9.4\\ s=1.7\\ N=150\end{array}$
4.	$P_u=0.57$	$\begin{array}{l}{P}_s=0.60\\ N=117\end{array}$
5.	$P_u=0.32$	$\begin{array}{l}{P}_s=0.30\\ N=322\end{array}$

To determine

a)

To find:

The Z(critical) value for the given situation.

Answer to Problem 7.1P

Solution:

The Z(critical) values are given in Table $\left(2\right)$.

Alpha ($\text{α}$)	Form of the Test	Z(critical)
0.05	One-tailed $\alpha =0.05$	$+1.645$ or $-1.645$
0.10	Two-tailed $\alpha /2=0.05$	$\pm 1.645$
0.06	Two-tailed $\alpha /2=0.03$	$\pm 1.881$
0.01	One-tailed $\alpha =0.01$	$+2.326$ or $-2.326$
0.02	Two-tailed $\alpha /2=0.01$	$\pm 2.326$

Explanation of Solution

Given:

The given table of alpha values is,

Alpha ($\text{α}$)	Form of the Test	Z(critical)
0.05	One-tailed
0.10	Two-tailed
0.06	Two-tailed
0.01	One-tailed
0.02	Two-tailed

Description:

The critical region of a sampling distribution is the area that includes the unlikely sample outcomes.

The critical Z values are the areas under the standard normal curve.

For one-tailed critical values, the area is only on one side and the whole value of $\alpha$ is taken.

For two-tailed critical values, the area is equally divided on both the sides and the value of $\alpha$ taken is $\alpha /2$.

Calculation:

The given table of alpha values is,

Alpha ($\text{α}$)	Form of the Test	Z(critical)
0.05	One-tailed
0.10	Two-tailed
0.06	Two-tailed
0.01	One-tailed
0.02	Two-tailed

Table $\left(1\right)$

The Z(critical) are obtained from the critical value table for standard normal variates.

For $\alpha =0.05$, the one-tailed Z(critical) values are given from the critical value table as,

$Z_{0.05}=-1.645$

Or,

$Z_{0.05}=+1.645$

Thus, for $\alpha =0.05$, the one-tailed Z(critical) values are $+1.645$ or $-1.645$.

For $\alpha =0.10$, the two-tailed Z(critical) values are obtained from the critical value table as,

$\begin{array}{rll}{Z}_{0.10/2}& =& Z_{0.05}\\ & =& -1.645\end{array}$

And,

$\begin{array}{rll}{Z}_{0.10/2}& =& Z_{0.05}\\ & =& +1.645\end{array}$

Thus, for $\alpha =0.10$, the two-tailed Z(critical) values are $\pm 1.645$.

For $\alpha =0.06$, the two-tailed Z(critical) values are obtained from the critical value table as,

$\begin{array}{rll}{Z}_{0.06/2}& =& Z_{0.03}\\ & =& -1.881\end{array}$

And,

$\begin{array}{rll}{Z}_{0.06/2}& =& Z_{0.03}\\ & =& +1.881\end{array}$

Thus, for $\alpha =0.06$, the two-tailed Z(critical) values are $\pm 1.881$.

For $\alpha =0.01$, the one-tailed Z(critical) values are obtained from the critical value table as,

$Z_{0.01}=-2.326$

Or,

$Z_{0.01}=+2.326$

Thus, for $\alpha =0.01$, the one-tailed Z(critical) values are $+2.326$ or $-2.326$.

For $\alpha =0.02$, the two-tailed Z(critical) values are obtained from the critical value table as,

$\begin{array}{rll}{Z}_{0.02/2}& =& Z_{0.01}\\ & =& -2.326\end{array}$

And,

$\begin{array}{rll}{Z}_{0.02/2}& =& Z_{0.01}\\ & =& +2.326\end{array}$

Thus, for $\alpha =0.02$, the two-tailed Z(critical) values are $\pm 2.326$.

The obtained critical values are tabulated as,

Alpha ($\text{α}$)	Form of the Test	Z(critical)
0.05	One-tailed $\alpha =0.05$	$+1.645$ or $-1.645$
0.10	Two-tailed $\alpha /2=0.05$	$\pm 1.645$
0.06	Two-tailed $\alpha /2=0.03$	$\pm 1.881$
0.01	One-tailed $\alpha =0.01$	$+2.326$ or $-2.326$
0.02	Two-tailed $\alpha /2=0.01$	$\pm 2.326$

Table $\left(2\right)$

Conclusion:

The Z(critical) values are given in Table $\left(2\right)$.

Alpha ($\text{α}$)	Form of the Test	Z(critical)
0.05	One-tailed $\alpha =0.05$	$+1.645$ or $-1.645$
0.10	Two-tailed $\alpha /2=0.05$	$\pm 1.645$
0.06	Two-tailed $\alpha /2=0.03$	$\pm 1.881$
0.01	One-tailed $\alpha =0.01$	$+2.326$ or $-2.326$
0.02	Two-tailed $\alpha /2=0.01$	$\pm 2.326$

To determine

b)

To find:

The t(critical) value for the given situation.

Answer to Problem 7.1P

Solution:

The t(critical) values are given in Table $\left(4\right)$.

Alpha ($\text{α}$)	Form of the Test	N	t(critical)
0.10	Two-tailed $\alpha /2=0.05$	31	$\pm 1.697$
0.02	Two-tailed $\alpha /2=0.01$	24	$\pm 2.5$
0.01	Two-tailed $\alpha /2=0.005$	121	$\pm 2.617$
0.01	One-tailed $\alpha =0.01$	31	$+2.457$ or $-2.457$
0.05	One-tailed $\alpha =0.05$	61	$+1.671$ or $-1.671$

Explanation of Solution

Given:

The given table of alpha values is,

Alpha ($\text{α}$)	Form of the Test	N	t(critical)
0.10	Two-tailed	31
0.02	Two-tailed	24
0.01	Two-tailed	121
0.01	One-tailed	31
0.05	One-tailed	61

Description:

The critical region of a sampling distribution is the area that includes the unlikely sample outcomes.

The critical t values are the areas under the curve of t-distribution.

For one-tailed critical values, the area is only on one side and the whole value of $\alpha$ is taken.

For two-tailed critical values, the area is equally divided on both the sides and the value of $\alpha$ taken is $\alpha /2$.

Calculation:

The given table of alpha values is,

Alpha ($\text{α}$)	Form of the Test	N	t(critical)
0.10	Two-tailed	31
0.02	Two-tailed	24
0.01	Two-tailed	121
0.01	One-tailed	31
0.05	One-tailed	61

Table $\left(3\right)$

The t(critical) values are obtained from the t-distribution table with the given value of $\alpha$ and $N-1$ degrees of freedom (df), where N is the sample size.s

For $\alpha =0.10$ and $N=31$, the two-tailed t(critical) values are obtained from the critical value table as,

$\begin{array}{rll}df& =& 31-1\\ & =& 30\end{array}$

$\begin{array}{rll}{t}_{\left(30,0.10/2\right)}& =& t_{\left(30,0.05\right)}\\ & =& -1.697\end{array}$

And,

$\begin{array}{rll}{t}_{\left(30,0.10/2\right)}& =& t_{\left(30,0.05\right)}\\ & =& +1.697\end{array}$

Thus, for $\alpha =0.10$ and $N=31$, the two-tailed t(critical) values are $\pm 1.697$.

For $\alpha =0.02$ and $N=24$, the two-tailed t(critical) values are obtained from the critical value table as,

$\begin{array}{rll}df& =& 24-1\\ & =& 23\end{array}$

$\begin{array}{rll}{t}_{\left(23,0.02/2\right)}& =& t_{\left(23,0.01\right)}\\ & =& -2.5\end{array}$

And,

$\begin{array}{rll}{t}_{\left(23,0.02/2\right)}& =& t_{\left(23,0.01\right)}\\ & =& +2.5\end{array}$

Thus, for $\alpha =0.02$ and $N=24$, the two-tailed t(critical) values are $\pm 2.5$.

For $\alpha =0.01$ and $N=121$, the two-tailed t(critical) values are obtained from the critical value table as,

$\begin{array}{rll}df& =& 121-1\\ & =& 120\end{array}$

$\begin{array}{rll}{t}_{\left(120,0.01/2\right)}& =& t_{\left(121,0.005\right)}\\ & =& -2.617\end{array}$

And,

$\begin{array}{rll}{t}_{\left(120,0.01/2\right)}& =& t_{\left(121,0.005\right)}\\ & =& +2.617\end{array}$

Thus, for $\alpha =0.01$ and $N=121$, the two-tailed t(critical) values are $\pm 2.617$.

For $\alpha =0.01$ and $N=31$, the one-tailed t(critical) values are obtained from the critical value table as,

$\begin{array}{rll}df& =& 31-1\\ & =& 30\end{array}$

$t_{\left(30,0.01\right)}=-2.457$

Or,

$t_{\left(30,0.01\right)}=+2.457$

Thus, for $\alpha =0.01$ and $N=31$, the two-tailed t(critical) value is $+2.457$ or $-2.457$.

For $\alpha =0.05$ and $N=61$, the one-tailed t(critical) values are obtained from the critical value table as,

$\begin{array}{rll}df& =& 61-1\\ & =& 60\end{array}$

$t_{\left(60,0.05\right)}=-1.671$

Or,

$t_{\left(60,0.05\right)}=+1.671$

Thus, for $\alpha =0.05$ and $N=61$, the two-tailed t(critical) value is $+1.671$ or $-1.671$.

The obtained critical values are tabulated as,

Alpha ($\text{α}$)	Form of the Test	N	t(critical)
0.10	Two-tailed $\alpha /2=0.05$	31	$\pm 1.697$
0.02	Two-tailed $\alpha /2=0.01$	24	$\pm 2.5$
0.01	Two-tailed $\alpha /2=0.005$	121	$\pm 2.617$
0.01	One-tailed $\alpha =0.01$	31	$+2.457$ or $-2.457$
0.05	One-tailed $\alpha =0.05$	61	$+1.671$ or $-1.671$

Table $\left(4\right)$

Conclusion:

The t(critical) values are given in Table $\left(4\right)$.

Alpha ($\text{α}$)	Form of the Test	N	t(critical)
0.10	Two-tailed $\alpha /2=0.05$	31	$\pm 1.697$
0.02	Two-tailed $\alpha /2=0.01$	24	$\pm 2.5$
0.01	Two-tailed $\alpha /2=0.005$	121	$\pm 2.617$
0.01	One-tailed $\alpha =0.01$	31	$+2.457$ or $-2.457$
0.05	One-tailed $\alpha =0.05$	61	$+1.671$ or $-1.671$

To determine

c)

To find:

The appropriate test statistics for the given situation.

Answer to Problem 7.1P

Solution:

The Z(critical) and t(critical) values are given in Table $\left(6\right)$.

	Population	Sample	Z(obtained) or t(obtained)
1.	$\begin{array}{l}\mu =2.40\\ \sigma =0.75\end{array}$	$\begin{array}{l}\overline{X}=0.20\\ N=200\end{array}$	$-3.77$
2.	$\mu =17.1$	$\begin{array}{l}\overline{X}=16.8\\ s=0.9\\ N=45\end{array}$	$-2.24$
3.	$\mu =10.2$	$\begin{array}{l}\overline{X}=9.4\\ s=1.7\\ N=150\end{array}$	$-5.76$
4.	$P_u=0.57$	$\begin{array}{l}{P}_s=0.60\\ N=117\end{array}$	0.66
5.	$P_u=0.32$	$\begin{array}{l}{P}_s=0.30\\ N=322\end{array}$	0.77

Explanation of Solution

Given:

The given table of information is,

	Population	Sample	Z(obtained) or t(obtained)
1.	$\begin{array}{l}\mu =2.40\\ \sigma =0.75\end{array}$	$\begin{array}{l}\overline{X}=0.20\\ N=200\end{array}$
2.	$\mu =17.1$	$\begin{array}{l}\overline{X}=16.8\\ s=0.9\\ N=45\end{array}$
3.	$\mu =10.2$	$\begin{array}{l}\overline{X}=9.4\\ s=1.7\\ N=150\end{array}$
4.	$P_u=0.57$	$\begin{array}{l}{P}_s=0.60\\ N=117\end{array}$
5.	$P_u=0.32$	$\begin{array}{l}{P}_s=0.30\\ N=322\end{array}$

Formula used:

For large samples with single mean and given standard deviation, the Z value is given by,

$Z\left(\text{obtained}\right)=\frac{\overline{X}-\mu }{\sigma /\sqrt{N}}$

Where, $\overline{X}$ is the sample mean,

$\mu$ is the population,

$\sigma$ is the population standard deviation and

N is the sample size.

For small samples with single sample mean and unknown standard deviation, the t value is given by,

$t\left(\text{obtained}\right)=\frac{\overline{X}-\mu }{s/\sqrt{N-1}}$

Where, $\overline{X}$ is the sample mean,

$\mu$ is the population mean,

$s$ is the estimate population standard deviation and,

N is the sample size.

For large samples with single sample proportions, the Z value is given by,

$Z\left(\text{obtained}\right)=\frac{P_s-P_u}{\sqrt{P_u\left(1-P_u\right)/N}}$

Where, $P_s$ is the sample proportion,

$P_u$ is the population proportion and,

N is the sample size.

Calculation:

The given table of alpha values is,

	Population	Sample	Z(obtained) or t(obtained)
1.	$\begin{array}{l}\mu =2.40\\ \sigma =0.75\end{array}$	$\begin{array}{l}\overline{X}=0.20\\ N=200\end{array}$
2.	$\mu =17.1$	$\begin{array}{l}\overline{X}=16.8\\ s=0.9\\ N=45\end{array}$
3.	$\mu =10.2$	$\begin{array}{l}\overline{X}=9.4\\ s=1.7\\ N=150\end{array}$
4.	$P_u=0.57$	$\begin{array}{l}{P}_s=0.60\\ N=117\end{array}$
5.	$P_u=0.32$	$\begin{array}{l}{P}_s=0.30\\ N=322\end{array}$

Table $\left(5\right)$

For large samples with single mean and given standard deviation, the Z value is given by,

$Z\left(\text{obtained}\right)=\frac{\overline{X}-\mu }{\sigma /\sqrt{N}}$ ……$\left(1\right)$

For sample mean $\overline{X}=0.20$, population mean $\mu =2.40$, population standard deviation $\sigma =0.75$ and sample size $N=200$, the Z value is given by,

Substitute 2.20 for $\overline{X}$, 2.40 for $\mu$, 0.75 for $\sigma$ and 200 for $N$ in equation(1).

$\begin{array}{rll}Z\left(\text{obtained}\right)& =& \frac{2.20-2.40}{0.75/\sqrt{200}}\\ & =& \frac{-0.20}{0.053}\\ & =& -3.77\end{array}$

Thus, the obtained Z value is $-3.77$.

For small samples with single sample mean and unknown standard deviation, the t value is given by,

$t\left(\text{obtained}\right)=\frac{\overline{X}-\mu }{s/\sqrt{N-1}}$ ……$\left(2\right)$

For sample mean $\overline{X}=16.8$, population mean $\mu =17.1$, sample standard deviation $s=0.9$ and sample size $N=45$, the t value is given by,

Substitute 16.8 for $\overline{X}$, 17.1 for $\mu$, 0.9 for $s$ and 45 for $N$ in equation(2).

$\begin{array}{rll}t\left(\text{obtained}\right)& =& \frac{16.8-17.1}{0.9/\sqrt{45}}\\ & =& \frac{-0.3}{0.1342}\\ & =& -2.24\end{array}$

Thus, the obtained t value is $-2.24$.

For sample mean $\overline{X}=9.4$, population mean $\mu =10.2$, sample standard deviation $s=1.7$ and sample size $N=150$, the t value is given by,

Substitute 9.4 for $\overline{X}$, 10.2 for $\mu$, 1.7 for $s$ and 150 for $N$ in equation(2).

$\begin{array}{rll}t\left(\text{obtained}\right)& =& \frac{9.4-10.2}{1.7/\sqrt{150}}\\ & =& \frac{-0.8}{0.1388}\\ & =& -5.76\end{array}$

Thus, the obtained t value is $-5.76$.

For large samples with single sample proportions, the Z value is given by,

$Z\left(\text{obtained}\right)=\frac{P_s-P_u}{\sqrt{P_u\left(1-P_u\right)/N}}$ ……$\left(3\right)$

For sample proportion $P_s=0.60$, population proportion $P_u=0.57$, and sample size $N=117$, the Z value is given by,

Substitute 0.60 for $P_s$, 0.57 for $P_u$ and 117 for $N$ in equation(3).

$\begin{array}{rll}Z\left(\text{obtained}\right)& =& \frac{0.6-0.57}{\sqrt{0.57\left(1-0.57\right)/117}}\\ & =& \frac{0.03}{0.0458}\\ & =& 0.66\end{array}$

Thus, the obtained Z value is $0.66$.

For sample proportion $P_s=0.30$, population proportion $P_u=0.32$, and sample size $N=322$, the Z value is given by,

Substitute 0.30 for $P_s$, 0.32 for $P_u$ and 322 for $N$ in equation(3).

$\begin{array}{rll}Z\left(\text{obtained}\right)& =& \frac{0.30-0.32}{\sqrt{0.32\left(1-0.32\right)/322}}\\ & =& \frac{-0.02}{0.026}\\ & =& 0.77\end{array}$

Thus, the obtained Z value is $0.77$.

The obtained critical values are tabulated as,

	Population	Sample	Z(obtained) or t(obtained)
1.	$\begin{array}{l}\mu =2.40\\ \sigma =0.75\end{array}$	$\begin{array}{l}\overline{X}=0.20\\ N=200\end{array}$	$-3.77$
2.	$\mu =17.1$	$\begin{array}{l}\overline{X}=16.8\\ s=0.9\\ N=45\end{array}$	$-2.24$
3.	$\mu =10.2$	$\begin{array}{l}\overline{X}=9.4\\ s=1.7\\ N=150\end{array}$	$-5.76$
4.	$P_u=0.57$	$\begin{array}{l}{P}_s=0.60\\ N=117\end{array}$	0.66
5.	$P_u=0.32$	$\begin{array}{l}{P}_s=0.30\\ N=322\end{array}$	0.77

Table $\left(6\right)$

Conclusion:

The Z(critical) and t(critical) values are given in Table $\left(6\right)$.

	Population	Sample	Z(obtained) or t(obtained)
1.	$\begin{array}{l}\mu =2.40\\ \sigma =0.75\end{array}$	$\begin{array}{l}\overline{X}=0.20\\ N=200\end{array}$	$-3.77$
2.	$\mu =17.1$	$\begin{array}{l}\overline{X}=16.8\\ s=0.9\\ N=45\end{array}$	$-2.24$
3.	$\mu =10.2$	$\begin{array}{l}\overline{X}=9.4\\ s=1.7\\ N=150\end{array}$	$-5.76$
4.	$P_u=0.57$	$\begin{array}{l}{P}_s=0.60\\ N=117\end{array}$	0.66
5.	$P_u=0.32$	$\begin{array}{l}{P}_s=0.30\\ N=322\end{array}$	0.77