Respuesta :
Answer:
Yes, this indicate that the percentage has increased from that of previous studies.
Step-by-step explanation:
We are given that Results from previous studies showed 76% of all high school seniors from a certain city plan to attend college after graduation.
A random sample of 200 high school seniors from this city reveals that 162 plan to attend college.
Let p = % of all high school seniors from a certain city who plan to attend college after graduation
SO, Null Hypothesis, [tex]H_0[/tex] : p [tex]\leq[/tex] 76% {means that the percentage has not increased from that of previous studies}
Alternate Hypothesis, [tex]H_a[/tex] : p > 76% {means that the percentage has increased from that of previous studies}
The test statistics that will be used here is One-sample z proportion statistics;
T.S. = [tex]\frac{\hat p-p}{{\sqrt{\frac{\hat p(1-\hat p)}{n} } } } }[/tex] ~ N(0,1)
where, [tex]\hat p[/tex] = sample proportion of high school seniors from this city who
plan to attend college = [tex]\frac{162}{200}[/tex] = 0.81
n = sample of high school seniors = 200
So, test statistics = [tex]\frac{0.81-0.76}{{\sqrt{\frac{0.81(1-0.81)}{200} } } } }[/tex]
= 1.8025
Now at 5% significance level, the z table gives critical value of 1.6449 for right-tailed test. Since our test statistics is more than the critical value of z so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.
Therefore, we conclude that the percentage has increased from that of previous studies.
