Chapter 9.4, Problem 9.18E

i.

To determine

Alternative and null hypothesis for the test.

Answer to Problem 9.18E

Null hypothesis, $H_0:{\mu }_1-{\mu }_2=0$

Alternative hypothesis, $H_{\alpha }:{\mu }_1-{\mu }_2>0$

Explanation of Solution

Given:

Given statement is: ${\mu }_1$ is the larger than the ${\mu }_2$ or ${\mu }_1-{\mu }_2>0$ .

1. No of sample, $n_1=80$
2. No of sample, $n_2=80$
3. ${\overline{x}}_1=$ Sample mean $=11.6$
4. $\overline{x_2}=$ Sample mean $=9.7$
5. Standard deviation, $s_1^2=27.9$
6. Standard deviation, $s_2^2=38.4$
7. $\alpha =$ Significance level $=1\%=0.01$ .

Calculation:

Given claim is either the null hypothesis or the alternative hypothesis. The null hypothesis states that the population mean is equal to the value mentioned in the given statement.

In case, null hypothesis is the claim, then the alternative hypothesis states the opposite of the null hypothesis.

Null hypothesis, $H_0:{\mu }_1-{\mu }_2=0$

Alternative hypothesis, $H_{\alpha }:{\mu }_1-{\mu }_2>0$

Conclusion:

Therefore, Null hypothesis, $H_0:{\mu }_1-{\mu }_2=0$ and Alternative hypothesis, $H_{\alpha }:{\mu }_1-{\mu }_2>0$

ii.

To determine

To identify: Given test is one-tail or two tail.

Answer to Problem 9.18E

Test is one-tailed.

Explanation of Solution

Given:

1. No of sample, $n_1=80$
2. No of sample, $n_2=80$
3. ${\overline{x}}_1=$ Sample mean $=11.6$
4. $\overline{x_2}=$ Sample mean $=9.7$
5. Standard deviation, $s_1^2=27.9$
6. Standard deviation, $s_2^2=38.4$
7. $\alpha =$ Significance level $=1\%=0.01$ .

Calculation:

Given claim is either the null hypothesis or the alternative hypothesis. The null hypothesis states that the population mean is equal to the value mentioned in the given statement.

In case, null hypothesis is the claim, then the alternative hypothesis states the opposite of the null hypothesis.

Null hypothesis, $H_0:{\mu }_1-{\mu }_2=0$

Alternative hypothesis, $H_{\alpha }:{\mu }_1-{\mu }_2>0$

If the alternative hypothesis contains is $(<)$ or $(>)$ then the test is one-tailed.

If the alternative hypothesis contains is $\ne$ the test is two-tailed.

Here analternative hypothesis is $H_{\alpha }:{\mu }_1-{\mu }_2>0$ thus, the given test is one-tailed.

Conclusion:

Therefore,as alternative hypothesis is $H_{\alpha }:{\mu }_1-{\mu }_2>0$ thus, test is one-tailed.

iii.

To determine

Test statistic for the part (i).

Answer to Problem 9.18E

  $z$-score is $2.09$data present is sufficient evidence to indicate that mean for population $1$and $2$.

Explanation of Solution

Given:

1. No of sample, $n_1=80$
2. No of sample, $n_2=80$
3. ${\overline{x}}_1=$ Sample mean $=11.6$
4. $\overline{x_2}=$ Sample mean $=9.7$
5. Standard deviation, $s_1^2=27.9$
6. Standard deviation, $s_2^2=38.4$
7. $\alpha =$ Significance level $=1\%=0.01$ .

Formula used:

Test statistic:

  $z=\frac{\overline{x}-\mu }{s/\sqrt{n}}$

Calculation:

Given claim is either the null hypothesis or the alternative hypothesis. The null hypothesis states that the population mean is equal to the value mentioned in the given statement.

In case, null hypothesis is the claim, then the alternative hypothesis states the opposite of the null hypothesis.

Null hypothesis, $H_0:{\mu }_1-{\mu }_2=0$

Alternative hypothesis, $H_{\alpha }:{\mu }_1-{\mu }_2>0$

Since the sample size $n_1$ and $n_2$ are large (at least $30$ ), it is appropriate to find out the population deviation by the sample standard deviation.

  $\begin{array}{l}{\sigma }_1\approx s_1\\ {\sigma }_2\approx s_2\end{array}$

Now, for finding out the test statistic:

  $\begin{array}{l}z=\frac{{\overline{x}}_1-{\overline{x}}_2-D_0}{\sqrt{\frac{{\sigma }_1^2}{n_1}}+\frac{{\sigma }_2^2}{n_2}}\\ =\frac{11.6-9.7}{\sqrt{\frac{27.9}{80}}+\frac{38.4}{80}}\\ \approx 2.09\end{array}$

A $z$ -score is considered unusual if it is less than $-2$ or larger than $2$ .

  $z$ -score is $2.09$ is larger than $2$ and thus, it is an unlikely observation. Do not reject $H_0$ at $1\%$ level of significance.

Conclusion:

Therefore, $z$ -score is $2.09$ data present is sufficient evidence to indicate that mean for population $1$ and $2$ .

So, the test is one-tailed.

iv.

To determine

The $P$ -value for the test.

Answer to Problem 9.18E

The $P$-value for the given test is $0.0183$and there is no sufficient evidence to conclude ${\mu }_1$is greater than the ${\mu }_2$.

Explanation of Solution

Given:

Given statement: Mean at difference in the population.

1. No of sample, $n_1=80$
2. No of sample, $n_2=80$
3. ${\overline{x}}_1=$ Sample mean $=11.6$
4. $\overline{x_2}=$ Sample mean $=9.7$
5. Standard deviation, $s_1^2=27.9$
6. Standard deviation, $s_2^2=38.4$
7. $\alpha =$ Significance level $=1\%=0.01$ .
8. $z=2.09$

Formula Used:

Test statistics:

  $z=\frac{\overline{x}-\mu }{s/\sqrt{n}}$

Calculation:

Given claim is either the null hypothesis or the alternative hypothesis. The null hypothesis states that the population mean is equal to the value mentioned in the given statement.

In case, null hypothesis is the claim, then the alternative hypothesis states the opposite of the null hypothesis.

Null hypothesis, $H_0:{\mu }_1-{\mu }_2=0$

Alternative hypothesis, $H_{\alpha }:{\mu }_1-{\mu }_2>0$

If the alternative hypothesis contains is $(<)$ then the test is left-tailed.

If the alternative hypothesis contains is $(>)$ then the test is right-tailed.

If the alternative hypothesis contains is $\ne$ the test is two-tailed.

The P-value is the probability of obtaining a value more extreme or equal to the standardized test statistic $z$ , assuming that the null hypothesis is true.

Since the test is right-tailed, the P-value is the probability that the $z$ score is greater than the $z=2.09$

  $P\left(z>2.09\right)$

Now, to find corresponding probability using the normal probability table. the $P\left(z<2.09\right)$ is lies in row starting with $1.1$ and in the column starting with the $0.05$ of table.

The rejection region for this two-tail test of the hypothesis is found in the both tails of normal probability distribution.

Since, the observed value of the test statistic is $2.09$ , $\alpha =0.05$ at level of significance is $1.96$ .

The $P$ -value for the given large sample $z$ test is,

  $\begin{array}{l}P\left(z>2.09\right)\\ =1-P\left(z>2.09\right)\\ =1-0.9817\\ =0.0183\end{array}$

The P-value if greater than the $\alpha$ , sothere is no sufficient evidence to conclude ${\mu }_1$ is greater than the ${\mu }_2$ .

Conclusion:

Therefore, the $P$ -value for the given test is $0.0183$ and there is no sufficient evidence to conclude ${\mu }_1$ is greater than the ${\mu }_2$ .

v.

To determine

To identify: data is sufficient to indicate the difference population mean and rejection region when $\alpha =0.01$ by critical value approach.

Answer to Problem 9.18E

There is no sufficient evidence to conclude that ${\mu }_1$is more than ${\mu }_2$.

Explanation of Solution

Given:

Given statement: Mean at difference in the population.

1. No of sample, $n_1=80$
2. No of sample, $n_2=80$
3. ${\overline{x}}_1=$ Sample mean $=11.6$
4. $\overline{x_2}=$ Sample mean $=9.7$
5. Standard deviation, $s_1^2=27.9$
6. Standard deviation, $s_2^2=38.4$
7. $\alpha =$ Significance level $=1\%=0.01$ .

Formula Used:

Complement rule:

  $P\left(A^c\right)=P\left(notA\right)=1-P\left(A\right)$

Calculation:

states that the population mean is equal to the value mentioned in the given statement.

In case, null hypothesis is the claim, then the alternative hypothesis states the opposite of the null hypothesis.

Null hypothesis, $H_0:{\mu }_1-{\mu }_2=0$

Alternative hypothesis, $H_{\alpha }:{\mu }_1-{\mu }_2>0$

If the alternative hypothesis contains is $(<)$ then the test is left-tailed.

If the alternative hypothesis contains is $(>)$ then the test is right-tailed.

If the alternative hypothesis contains is $\ne$ the test is two-tailed.

The rejection region of right-tailed test with $\alpha =0.01$ contains all $z-$ scores above the $z-$ scores $z_0$ that has a probability of $0.01$ to the right side.

  $P\left(z>z_0\right)=0.01$

By using Complement rule:

  $\begin{array}{l}P\left(A^c\right)=P\left(notA\right)=1-P\left(A\right)\\ P\left(z<z_0\right)=1-P\left(z>z_0\right)=1-0.01=0.99\end{array}$

Now, to find the $z$ -score with corresponding probability $0.99$ in normal probability table.

The probability $0.99$ lies in row starting with $2.3$ and in the column starting with the $0.03$ of table.

  $z_0=2.33$

The rejection region then contains all $z$ -score more than $2.33$ .

If the $z$ score $2.33$ is in the rejection region then the null hypothesis is rejected.

  $2.09<2.33$

Failed to reject $H_0$ .

Conclusion:

Therefore, there is no sufficient evidence to conclude that ${\mu }_1$ is more than ${\mu }_2$ .