91影视

Referring to Exercises 8.8 and 8.9, there is no difference between the n1 and n2 consumers' average scores. The sample mean of consumers with a flexed arm and extended arm was different, that is$\stackrel{炉}{x}1$= 59,$\stackrel{炉}{x}2$= 43. The standard deviation in consumers with a flexed arm and consumers with an extended are dissimilar.

A statistical test called a t-test is employed to contrast the means of two clusters. It is frequently employed in hypothesis testing to establish whether one procedure or treatment affects the target group or even if two groups vary.

The two different kinds of t-tests are as follows.

• Use a two-tailed t-test if the only thing that matters is how the two populations vary from each other.
• Use a one-tailed t-test to determine if one population mean is higher or lower compared to the other.

t-test hypotheses are as follows

$$\begin{gathered} H0:渭1=渭2 \\ H1:渭1鈮�渭2 \end{gathered}$$

The application of test statistics is as follows.

$t=\frac{\left(\stackrel{炉}{x}1-\stackrel{炉}{x}2\right)-\left(渭1^{}-渭2\right)}{\sqrt{\frac{S{1}^2}{n1}+\frac{S{2}^2}{n2}}}$

Following is test statistics with the degree of freedom.

$df=n1-1$and $df=n_2-1$are similar because of the mean of consumers with a flexed arm and extended arm are similar. As $H_0:渭1=渭2$and $n_1=n_2$, reject the claim that the value of t is higher. Therefore, the hypothesis will most likely be rejected if $S_1$and $S_2$ are lower.

Case-1:If $S_1=4$and $S_2=2$

The test statistics is as follows.

$$\begin{gathered} t=\frac{59-43}{\sqrt{\frac{4^2}{11}+\frac{2^2}{11}}} \\ t=\frac{16}{\sqrt{\frac{20}{11}}} \\ t=\sqrt[8]{\frac{11}{5}} \\ t=11.866 \end{gathered}$$

Case-2:If $S_1=10$and $S_2=15$

$$\begin{gathered} t=\frac{59-43}{\sqrt{\frac{{10}^2}{11}+\frac{{15}^2}{11}}} \\ t=\sqrt[16]{\frac{11}{325}} \\ t=2.946 \end{gathered}$$

The p-value for case 1 will be higher than case 2 as t increases. In case 1, it is more likely that the hypotheses will be rejected.