Question

Research at the University of Toledo indicates that 50% of students change their major area of study after their first year in a program. A random sample of 100 students in the College of Business revealed that 48 had changed their major area of study after their first year of the program. Has there been a significant decrease in the proportion of students who change their major after the first year in this program? Test at the 0.05 level of significance.

Answer 1

`z=\frac{0.48 -0.5}{\sqrt{(0.5(1-0.5))/(100)}}=-0.4`

`p_v =P(z<-0.4)=0.345`

So the p value obtained was a very high value and using the significance level given
`\alpha=0.05` we have
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of students that had changed their major area of study after their first year of the program is not significantly less than 0.5 or 50%.

Explanation:

1) Data given and notation

n=100 represent the random sample taken

X=48 represent the students that had changed their major area of study after their first year of the program

`\hat p=(48)/(100)=0.48` estimated proportion of students that had changed their major area of study after their first year of the program

`p_o=0.5` is the value that we want to test

`\alpha=0.05` represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

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

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that if we have a significant decrease in the proportion of students who change their major after the first year in this program.:

Null hypothesis:
`p \geq 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{(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`.

3) Calculate the statistic

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

`z=\frac{0.48 -0.5}{\sqrt{(0.5(1-0.5))/(100)}}=-0.4`

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

Since is a left taild test the p value would be:

`p_v =P(z<-0.4)=0.345`

So the p value obtained was a very high value and using the significance level given
`\alpha=0.05` we have
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of students that had changed their major area of study after their first year of the program is not significantly less than 0.5 or 50%.