Answer:
A) Alternative hypothesis; Ha: p_o > ⅔
B) Null hypothesis;
H0: p_o = ⅔
C) There is sufficient evidence to support the claim that the community screening programme was effective
D) p-value = 0
Its less than the significance value and so we will reject the null hypothesis and conclude that the community screening programme was effective
Step-by-step explanation:
A) We are told that ⅓ of those diagnosed eventually die of the disease. Thus, ⅔ is the fraction that will survive the disease.
Thus, the null hypothesis is;
H0: p_o = ⅔
We are told that a screening programme was started to increase the survival rate. Thus,the alternative hypothesis is;
Ha: p_o > ⅔
B) null hypothesis is;
H0: p_o = ⅔
C) 164 out of the 200 women selected survived the disease.
Thus sample proportion is;
p^ = 164/200
p^ = 0.82
For the z-score value, we will use the formula;
z = (p^ - p_o)/√((p_o(1 - p_o)/n)
Now, p_o = ⅔ = 0.67
Thus;
z = (0.82 - 0.67)/√((0.67(1 - 0.67)/200)
z = 4.51
From online p-value from z-score calculator, we have p-value ≈ 0
This is less than the significant level of 0.05 and therefore we will reject the null hypothesis and conclude that the community screening program was effective.
D) p-value ≈ 0
Reject the null hypothesis
that simplifys down to 
and because fraction simplifications are similar (related to each other) we can concur that these would be too.