Question

Suppose that 34% of the students at a university favor the semester system, 61% favor the quarter system, and 5% have no preference. Is a random sample of 100 students large enough to provide convincing evidence that the quarter system is favored

Answer 1

Answer:

$z=\frac{0.61 -0.5}{\sqrt{\frac{0.5(1-0.5)}{100}}}=2.2$  

$p_v =P(z>2.2)=0.014$  

So the p value obtained was a very low value and using the significance level assumed $\alpha=0.05$ 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 favor the quarter system is higher than 0.5

Step-by-step explanation:

Data given and notation

n=100 represent the random sample taken

$\hat p=0.61$ estimated proportion of interest

$p_o=0.7$ is the value that we want to test

$\alpha$ represent the significance level

z would represent the statistic (variable of interest)

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

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion for quarter system is higher than 0.5.:  

Null hypothesis:$p \leq 0.5$  

Alternative hypothesis:$p > 0.5$  

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 from a hypothesized value $p_o$.

Calculate the statistic  

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

$z=\frac{0.61 -0.5}{\sqrt{\frac{0.5(1-0.5)}{100}}}=2.2$  

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 assumed is $\alpha=0.05$. The next step would be calculate the p value for this test.  

Since is a right tailedl test the p value would be:  

$p_v =P(z>2.2)=0.014$  

So the p value obtained was a very low value and using the significance level assumed $\alpha=0.05$ 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 favor the quarter system is higher than 0.5