Let p represents the population proportion.
By considering the given information, we have
[tex]H_0: p=0.29\\\\H_a: p\neq0.29[/tex]
∵ the alternative hypothesis is two tailed , so the test is two-tailed test.
Given : For sample size :n= 130, [tex]\hat{p}=0.23[/tex]
Test statistic: [tex]z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}[/tex]
[tex]z=\dfrac{0.23-0.29}{\sqrt{\dfrac{0.29(1-0.29)}{130}}}\\\\=-1.50762993902\approx-1.51[/tex]
P-value (two -tailed test)=[tex]2P(z>|-1.51|)=0.1310434\approx 0.131[/tex]
Since , the p-value (0.131) is greater than the significance level (0.03), so we accept the null hypothesis.
Thus , we conclude that we have sufficient evidence to support the null hypothesis that the true population proportion of Maryland students who have smoked marijuana is 29% .
