Question 647381
We test H0: p = 0.20 vs Ha: p < 0.20 taking alpha = 10% (ie. the claim of candidate Rita)
Let: P^ be the proportion in the survey
     Po be the proportion in the hypothesis
P^ = 180/1000 = 0.18 and Po = 0.20

Checking for 3 methods - confidence statements, critical boundaries and prob-value.


Method 1: Confidence statements
we test at the 90% cub for the proportion.
{{{P < P + Zalpha * (sqrt(P*(1-P))/sqrt(n))}}} Note the "P" after the < sign are P^ (ie P < P^ + Zalpha * sqrt(P^*(1-P^))/sqrt(n)
=> {{{P < 0.18 + invnorm(0.90) * (sqrt(0.18*(1-0.18))/sqrt(1000))}}}
{{{P < 0.18 + 1.28 * (sqrt(0.1476)/sqrt(1000))}}}
{{{P < 0.18 + 1.28 * 0.012149074}}}
{{{P < 0.18 + 0.0155696649}}}
P < 0.1956
Since, Po = 0.20 is not inside the confidence upper bound (P < 0.1956), we reject the null hypothesis (H0)- Rita's claim.


Method 2: Critical Boundaries
We test for the critical lower bound
{{{C = Po - Zalpha * (sqrt(Po(1-Po))/sqrt(n))}}}
{{{C = 0.20 - invnorm(0.90) * (sqrt(0.20*(1-0.20))/sqrt(1000))}}}
{{{C = 0.20 - 1.28 * (sqrt(0.16)/sqrt(1000))}}}
{{{C = 0.20 - 1.28 * 0.0126491106}}}
{{{C = 0.20 - 0.0161908616}}}
C = 0.1838
Since P^ = 0.18 is less than the critical lower bound (0.1838), we reject the null hypothesis (H0) - Rita's claim.


Method 3: The prob-value
We test the probability: P(P^<0.18)
=> {{{(P-Po)/sqrt(Po(1-Po))/sqrt(n) < (0.18-0.20)/sqrt(0.20*(1-0.20))/sqrt(1000))}}}
{{{Z < (0.18-0.20)/sqrt(0.20*(1-0.20))/sqrt(1000))}}}
{{{Z < (-0.020)/sqrt(0.16)/sqrt(1000))}}}
{{{Z < (-0.020)/0.0126491106}}}
{{{Z < -1.58113883}}}
Then, normalcdf(-10,-1.58113883) with the help of "graphical calculator" or "normal distribution table"
normalcdf(-10,-1.58113883) = 0.0569
Since the prob-value, 0.0569 or 5.6% is less than alpha of 10%, we reject the null hypothesis (H0) - Rita's claim.


In all, we reject candidate Rita's claim that 20% of electorates support her. 
Note: Either method applied, confidence or critical or prob-value, should give you the same conclusion - whether to reject or don't reject the H0.