Question

In a sample of 170 students at an Australian university that introduced the use of plagiarism-detection software in a number of courses, 58 students indicated a belief that such software unfairly targets students. Does this suggest that a majority of students at the university do not share this belief? Test appropriate hypotheses at level 0.05. (Let p be the proportion of students at this university who do not share this belief.)

Answer 1

`p_v =P(Z>4.146)=0.0000169`

Based on the p value obtained and using the significance level given
`\alpha=0.05` we have
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of students who NOT belief that such software unfairly targets students is higher than 0.5 or 50% .

Explanation:

1) Data given and notation

n=170 represent the random sample taken

X=58 represent the student's who belief that such software unfairly targets students

`\hat px=(58)/(170)=0.341` estimated proportion of students who belief that such software unfairly targets students

`\hat p=(112)/(170)=0.659` estimated proportion of students who NOT belief that such software unfairly targets students

`p_o=0.50` is the value that we want to test

`\alpha=0.05` represent the significance level (no given)

z would represent the statistic (variable of interest)

`p_v` represent the p value (variable of interest)

p= proportion of student's who belief that such software unfairly targets students

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that that a majority of students at the university do not share this belief. :

Null hypothesis:
`p\leq 0.5`

Alternative hypothesis:
`p>0.5`

We assume that the proportion follows a normal distribution.

When we conduct a proportion test we need to use the z statistic, and the is given by:

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

The One-Sample Proportion Test is used to assess whether a population proportion
`\hat p` is significantly (different,higher or less) from a hypothesized value
`p_o`.

Check for the assumptions that he sample must satisfy in order to apply the test

a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.

b) The sample needs to be large enough

`np_o =170*0.5=85>10`

`n(1-p_o)=170*(1-0.5)=85>10`

3) Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

`z=\frac{0.659 -0.5}{\sqrt{(0.5(1-0.5))/(170)}}=4.146`

4) Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided is
`\alpha=0.05`. The next step would be calculate the p value for this test.

Since is a one right side test the p value would be:

`p_v =P(Z>4.146)=0.0000169`

Based on the p value obtained and using the significance level given
`\alpha=0.05` we have
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of students who NOT belief that such software unfairly targets students is higher than 0.5 or 50% .