Chapter 10.2, Problem 47E

(a)

To determine

To Explain: Assign the day and used the partial line of random digits given below for Design A and Design B. Then use the plan to pick the first three days for Design A to be used.

Answer to Problem 47E

Day 21,24 and 55 were chosen for design A.

Explanation of Solution

Given:

every day of the 60 days a different number between 1 and 60.

chose the 1^st 2-digit number.

If the number is between 01 and 60, pick the corresponding day for design A, then forget the number and move on to the next 2-digit number. Repeat the process until design A has chosen 30 days and repeated numbers are ignored. To build A, the remaining 30 days are assigned.

find the 1^st 3 days for the design A:

  $\begin{array}{l}24\Rightarrow \text{Select day 24}\\ \text{00}\Rightarrow \text{Ignore}\\ \text{55}\Rightarrow \text{Select day 55}\\ \text{21}\Rightarrow \text{Select day 21}\end{array}$

Thus day 21, 21 and 55 were selected for design A.

(b)

To determine

To Explain: the choice that is it one-sided or two-sided significance test.

Answer to Problem 47E

Two-sided

  $\begin{array}{l}{H}_0:{\mu }_1={\mu }_2\\ H_a:{\mu }_1\ne {\mu }_2\end{array}$

Explanation of Solution

If there is a difference, testing is needed and then a two-sided significance test is best used. The null hypothesis says the mean of the population is equal:

  $H_0:{\mu }_1={\mu }_2$

The opposite of the null hypothesis is stated by the alternative hypothesis and uses the symbol since use the two-sided significance test.

  $H_a:{\mu }_1\ne {\mu }_2$

(c)

To determine

To Calculate: The P-value and degree of freedom.

Answer to Problem 47E

29

Explanation of Solution

Given:

  $\begin{array}{l}{n}_1=30\\ n_2=30\end{array}$

The degrees of freedom are

  $df=\text{min}\left(n_1-1,n_2-1\right)=\min \left(30-1,30-1\right)=29$

  $df=29$

(d)

To determine

To find: The P-value using the Table B and the conclusion.

Answer to Problem 47E

0.05

Explanation of Solution

Given:

  $\begin{array}{l}{n}_1=30\\ n_2=30\\ t=2.06\\ H_0:{\mu }_1={\mu }_2\left(\text{using part b}\right)\\ H_a:{\mu }_1\ne {\mu }_2\left(\text{using part b}\right)\end{array}$

Calculation:

The degrees of freedom are

  $df=\text{min}\left(n_1-1,n_2-1\right)=\min \left(30-1,30-1\right)=29$

The P-value is the

  $0.04=2\times 0.02<P<2\times 0.025=0.05$

  $P<0.05\Rightarrow \text{Reject }{H}_0$

There is enough proof to support the claim that there is a difference.