Sample proportion

Ok. So now to the hypothesis testing.
The standard deviation would be the square root of (.554*.446)/100 which is .0497.
The mean would be .554 the percentage given by answers.com.
To calculate the z-score   (.42-.554)/.0497= -2.696 z-scores away from the mean.
When converted to a percentage it is 0.351% and I am using a 5% significance so I must reject the null hypothesis that 55.4% of high school students have a job and favor that the true proportion of jobs is not 55.4%.

Testing Testing

The null hypothesis I am testing will be the 55.4%. The alternate hypothesis would be that the proportion of students that do not have a job would be not equal to 55.4%. I will use a 5% significance for this experiment to decide whether to reject the hypothesis or not.

Ok. So I finally have finished surveying 100 of my classmates.
The results...

• 42 out of 100 people have a job

 Now to the hypothesis testing!

Still Sampling...

I was busy with work in my classes yesterday so today I officially start sampling my classmates on wither or not they have a job. I found that I am making an assumption that students in the different levels (4.0, 5.0, and 6.0) have the same chance at getting a job because I am in mainly 6.0 classes.

Today I start my graduation project of AP Statistics. My topic like the name of the blog is about how many high school students have a job. I will be comparing the proportion of students that have a job that I find to the percentage given by Answers.com which is 55.4%.

I am going to use a cluster sampling method by asking students in my classes whether or not they have a job.