Answer)
Null hypothesis Ho : P <= 0.2 (20%)
Alternate hypothesis Ha : P> 0.2 (20%)
N = 270
First we need to check the conditions of normality that is if n*p and n*(1-p) both are greater than 5 or not
N*p = 54
N*(1-p) = 226
Both the conditions are met so we can use standard normal z table to estimate the P-Value
So type of test statistics is z test statistics for one sample proportion
Test statistics z = (oberved p - claimed p)/standard error
Standard error = √{claimed p*(1-claimed p)/√n
Observed P = 71/270
Claimed P = 0.2
N = 270
After substitution
Test statistics z = 2.59
From z table, P(z>2.59) = 0.0048
P-Value = 0.0048
As the p-value is less than 0.01(given significance level)
So we reject the null hypothesis Ho
So answer is yes
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