random variables and distributions

Comments

• Beau

  July 15 edited July 15

  Hi Jordan,
  I approached this as a one sample proportion test using a normal distribution to estamate probability( Z tables)

  we are given a P = 0.20 and a P-hat = $(20/60)$ = 0.33

  calculate your test statistic using $(P hat - P)/ SE $

  SE = $sqrt((P(1-P)/n)$
  Take this value to the Z table and find the p-value.
  Since the question states "more than 20 of the sixty" , you will take the value of
  (1-pvalue).
  I hope I did this correctly.

  
• mark

  July 15

  @Beau said:
  Hi Jordan,
  I approached this as a one sample proportion test using a normal distribution to estamate probability( Z tables)

  Yes, this is exactly right!

  The estimate $\hat{p}$ is a random variable whose mean and standard deviation are based on the actual proportion $p=0.2$ and the sample size $n=60$. More precisely, the mean of $\hat{p}$ is $np$ and the standard deviation of $\hat{p}$ is
  $\sqrt{p(1-p)/n}$. Thus, you use a normal distribution with those values for mean and standard deviation to do your estimate.