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FinanceConsultant

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The U.S. Department of Education reports that 46% of full-time college students are employed while attending college. (Data extracted from “The Condition of Education 2009,” National Center for Education Statistics, nces.ed. gov.) A recent survey of 60 full-time students at Miami University found that 29 were employed.
a. Use the five-step p-value approach to hypothesis testing and a 0.05 level of significance to determine whether the proportion of full-time students at Miami University is different from the national norm of 0.46.
b. Assume that the study found that 36 of the 60 full-time students were employed and repeat (a). Are the conclusions the same?
 
For you to learn this stuff, it’s probably better for you to work out the answers rather than merely to read what others here write about it. So, with that in mind:
  1. Is this a 1-tail test, or a 2-tail test?
  2. Is this a small sample or a large sample?
  3. Will you be using a Z-table or a t-table?
  4. If a t-table, what is the number of degrees of freedom?
  5. What are the critical values from the table?
Answer those and we’ll continue.
 
  1. Is this a 1-tail test, or a 2-tail test? 1 tail, different than norm?
  2. Is this a small sample or a large sample? Large, greater than 30.
  3. Will you be using a Z-table or a t-table? T, we do not have std deviation.
  4. If a t-table, what is the number of degrees of freedom? 60-1 = 59.
  5. What are the critical values from the table? .05
Guide me please.
 
So far, so good.
You’re misunderstanding question #5: what is the Z-value from the table?
 
FinanceConsultant wrote:No standard deviation. Therefore we must use the t table.
Click to expand...
Do you have a sample standard deviation?
(Sounds like a binomial distribution to me. What’s σ (or s)?)
 
But we need to calculate t-stat for our problem and we need standard deviation for the same.. what say?
 
FinanceConsultant wrote:I give up…
Click to expand...
Don’t give up just yet.
It’s a binomial distribution (success or failure: get a job, dont’ get a job). What’s the standard deviation for a binomial distribution?
 
^ I don’t know. I got the lightbulb problem. Thank you. Can you provide a walkthrough for this one? Spinning my tires here…
 
As I say, this is a binomial distribution: success or failure. Let’s call “employed while attending school” success, “unemployed while attending school” failure.
Our estimate of p – the probability of success – is k / n, where:
  • k = # of successes
  • n = sample size
So p = k / n = 29 / 60 = 0.4833.
For a binomial distribution, μ = n × p = 60 × 0.4833 = 29 (whew!), and σ = n × p × (1 – p) = 60 × 0.4833 × 0.5167 = 14.98.
Take it from there.
(By the way, the “p” I just calculated – the probability of success – isn’t the “p” about which they’re asking in this question. That’s statisticians for you.)
 
edupristine wrote:Need standard deviation to solve this. Missing
Click to expand...
It’s a binomial distribution; the sample standard deviation can be calculated from the sample size and the probability of success.
 
Before even attempting to calculate anything in the questions you’re posting (or getting other people to give you the answers) ask yourself this:
Do I understand what the p-value is, and what it means as a concept? Do I understand what the significance level is and what it means as a concept?
If you can’t answer these questions you’re not going to understand how to do any of these problems.
 
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