Question 1193687
Ho: <=23% of peas are 
Ha: >23% of peas are yellow
alpha=0.05 for this one way test prob {reject Ho|Ho true}
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normal approximation has mean np=136.62
variance is np(1-p)=105.20
sd is sqrt (V)=10.26
found 154
so z=(153.5-136.62)/10.256 using the continuity correction factor.
=1.645
probability z is >1.645 is 0.050. This is a one way test, so will be half the p-value of a two way. 

The answers to the multiple choice questions are A and B. One rejects Ho and concludes that there is sufficient evidence to warrant the claim that 23% of the peas are not yellow. 


The way I would do this is with a one-sample two way proportion test
Ho: p=0.23
Ha: P NE 0.23
alpha-0.05 p{reject Ho|Ho true}
test statistic is a z=(p hat-p)/sqrt(0.23*0.77/594)
critical value is |z|>1.96
p hat=0.02593
z= 0.0293/0.0173=1.693
fail to reject Ho, p-value is 0.090. Two-way tests double the p-value of one way, since both sides have areas in the rejection region.  This is more exact.