2

enter image description here

enter image description here

enter image description here

Was never really taught the use and interpretation of sample proportions explicitly. I tried to work through the first part of this problem above. Did I go wrong at any point?

aero26
  • 599

1 Answers1

0

Looks good to me. You could have obtained the sample proportion by taking

$$\hat{p} = 0.47\cdot0.57+0.53\cdot0.42=0.4905$$

If we set

$$H_0 : p=0.488$$

$$H_a : p \neq 0.488$$

We would fail to reject the null hypothesis since $0.488\in (0.4716,0.5084)$

Quick check of your calculations in R:

p = .47*.57+.53*.42
p+c(-1,1)*qnorm(1-.05/2)*sqrt(p*(1-p)/2935)

0.4724143 0.5085857

which is close to your confidence interval and is for all intents and purposes the same, except you rounded and used $z=2$ rather than $z=1.96$.

Remy
  • 8,128