Let p be the population proportion.
Then, Null hypothesis : [tex]H_0 : p=0.79[/tex]
Alternative hypothesis : [tex]H_a : p>0.79[/tex]
Since the alternative hypothesis is right tailed , then the hypothesis test is a right tailed test.
Given : A random sample of 200 high school seniors from this city reveals that 162 plan to attend college.
i.e. n = 200 > 30 , so we use z-test (otherwise we use t-test.)
[tex]\hat{p}=\dfrac{162}{200}=0.81[/tex]
Test statistic for population proportion :
[tex]z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}}[/tex]
[tex]=\dfrac{0.81-0.79}{\sqrt{\dfrac{0.81(1-0.81)}{200}}}=0.720984093915\approx0.72[/tex]
Thus, the z-value of the sample test-statistic = 0.72
P-value for right tailed test =[tex]P(z>0.72)=1-P(\leq0.72)[/tex]
[tex]=1-0.7642375=0.2357625\approx0.2358[/tex]
Since the p-value is greater than the significance level (0.05) , so we fail to reject the null hypothesis.
Hence, we conclude that we do not have sufficient evidence to support the claim that the percentage has increased from that of previous studies.
