Question

(Sec. 8.4) In a sample of 165 students at an Australian university that introduced the use of plagiarism-detection software in a number of courses, 55 students indicated a belief that such software unfairly targets students. Does this suggest that a majority of students at the university do not believe that it unfairly targets them? Test the appropriate hypotheses at the 5% significance level.

Answer 1

Answer:

$z=\frac{0.667 -0.5}{\sqrt{\frac{0.5(1-0.5)}{165}}}=4.29$  

The p value for this case is given by:

$p_v =P(z>4.29)=8.93x10^{-6}$  

Since the p value is lower than the significance level we have enough evidence to conclude that the true proportion is significantly higher than 0.5 at 5% of significance.

Step-by-step explanation:

Information given

n=165 represent the random sample selected

55 represent the students indicated a belief that such software unfairly targets students

X =165-55= 110 represent students who NOT belief that such software unfairly targets students

$\hat p=\frac{110}{165}=0.667$ estimated proportion of  students who NOT belief that such software unfairly targets students

$p_o=0.5$ is the value that we want to check

$\alpha=0.05$ represent the significance level

z would represent the statistic

$p_v$ represent the p value

System of hypothesis

We want to check if the majority of students at the university do not believe that it unfairly targets them, and the system of hypothesis are:

Null hypothesis:$p\leq 0.5$  

Alternative hypothesis:$p > 0.5$  

The statistic for this case is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)  

After replace we got:

$z=\frac{0.667 -0.5}{\sqrt{\frac{0.5(1-0.5)}{165}}}=4.29$  

The p value for this case is given by:

$p_v =P(z>4.29)=8.93x10^{-6}$  

Since the p value is lower than the significance level we have enough evidence to conclude that the true proportion is significantly higher than 0.5 at 5% of significance.