91Ó°ÊÓ

Given in the question that every day of the $60$days a different number between$1and60$. we have to describe how you would assign the days for Design A and Design B using the partial line of random digits provided below. Then use plan to select the first three days for using Design A.

choose the first two-digit number in the $1^{st}$text

If the number falls between $1and60$, choose the day for design A, then disregard the number and move on to the next $2$-digit number. Repeat the process until design A has selected $30$days, with repeated numbers being ignored. The remaining $30$days will be used to construct A.

for the design A, locate the $1^{st}$$3$days

$24⇒selectday24$

$00⇒ignore$

$55⇒selectday55$

$21⇒selectday21$

therefore,$21,24\&55$ were selected for design A

We have to determine that a one-sided or a two-sided significance test for this problem and Then set up appropriate hypothesis.

If a difference exists, testing is required, and a two-sided significance test is recommended. The null hypothesis states that the population's mean is the same:

$H_0:{\mathrm{μ}}_1={\mathrm{μ}}_2$

The alternative hypothesis states the opposite of the null hypothesis and utilizes the symbol since the two-sided significance test is used.

$H_{\mathrm{Î\pm }}:{\mathrm{μ}}_1≠{\mathrm{μ}}_2$

Given in the question that

$n_1=30$

$n_2=30$

we have to determine the degree of freedom.

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

Given in the question that

$n_1=30$

$n_2=30$

$t=2.06$

$H_0:{\mathrm{μ}}_1={\mathrm{μ}}_2$

$H_a:{\mathrm{μ}}_1≠{\mathrm{μ}}_2$

we have to find out the P value.

The degree of freedom is $29$

The P value is

$$\begin{gathered} 0.04=2×0.02<P<2×0.025 \\ =0.05 \end{gathered}$$

$P<0.05⇒reject{H}_0$

There is sufficient evidence to support the notion that there is a difference.