91影视

The model is trying to explain the likelihood that the client-provided explanation accounted for the fluctuation in the financial statement where the independent variables are client credibility (x₁) and linguistic delivery style (x₂) . In the model, variables x₁ and x₂ are said to have some interaction amongst them indicating that there is a relationship between the two variables which means that the client鈥檚 credibility might be related to the linguistic delivery style the client had chosen. This dependency is expressed using the term 鈥榵₁ x₂ 鈥�.

To check the overall goodness of the fit, the null hypothesis is whether the model parameters are explaining the model where the beta values are zero and the alternate hypothesis is whether the beta values are non-zero.

Mathematically,

$H_0:{尾}_1={尾}_2={尾}_3=0$

H_a At least one of the parameters ${尾}_1,{尾}_2,{尾}_3$is non zero

$H_0:{尾}_1={尾}_2={尾}_3=0$

At least one of the parameters${尾}_1,{尾}_2,{尾}_3$is non zero

Here, F test statistic =$\frac{SSE}{n-\left(k+1\right)}=55.35$

$H_0$is rejected if P - value < 0.01. For , since P - value < 0.0005

Sufficient evidence to reject $H_0$ at 95% confidence interval.

Therefore,${尾}_1鈮�{尾}_2鈮�{尾}_3鈮�0$

To test whether client credibility and linguistic delivery style interact, the value of ${尾}_3$ is tested

Mathematically,

localid="1651179732490" $$\begin{gathered} H_0:{尾}_3=0 \\ {H}_a:{尾}_3鈮�0 \end{gathered}$$

$$\begin{gathered} H_0:{尾}_3=0 \\ {H}_a:{尾}_3鈮�0 \end{gathered}$$

Here, t-test statistic $\frac{\stackrel{铃�}{{尾}_3}}{s{尾}_3}=\frac{0.036}{0.009}$

Value of $t_{0.05,199}$ is 1.645

is rejected if t static$>t_{0.05,199}$ . For$伪=0.05$, since t > $t_{0.05,199}$

Sufficient evidence to reject H₀ at 95% confidence interval.

Therefore,${尾}_3鈮�0$. Hence it can be concluded with enough evidence thatx₁and x₂do not interact in the model.

Given, $E\left(y\right)=15.865+0.037x_1-0.0678x_2+0.036x_1{x}_2$for x₁=22

localid="1651181050324" $$\begin{gathered} E\left(y\right)=15.865+0.037\left(22\right)-0.0678x_2+0.036\left(22\right)x_2 \\ E\left(y\right)=16.679+1.47x_2 \end{gathered}$$

The slope of the line relating y tox₂whenx₁= 22 is 1.47. The positive value denotes a positive relationship amongst the two variables and a low value means that the relation is not so strong.

Given, $E\left(y\right)=15.865+0.037x_1-0.0678x_2+0.036x_1{x}_2$ for x₁=46

$$\begin{gathered} E\left(y\right)=15.865+0.037\left(46\right)-0.0678x_2+0.036\left(46\right)x_2 \\ E\left(y\right)=17.567+2.334x_2 \end{gathered}$$

The slope of the line relating y tox₂whenx₁= 46 is 2.334. The positive value denotes a positive relationship amongst the two variables and a low value means that the relation is not so strong.