Based on the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v$ 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% .

Step-by-step 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

$\hat px=\frac{58}{170}=0.341$ estimated proportion of students who belief that such software unfairly targets students

$\hat p=\frac{112}{170}=0.659$ 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:

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

<em>Check for the assumptions that he sample must satisfy in order to apply the test </em>

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:

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

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:

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

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