Answer:
Do not reject [tex]H_0[/tex]. There is not enough evidence to support the claim that the proportion of students planning to go to college is greater than 0.79.
Step-by-step explanation:
The random sample reveals that 162/200 = 0.81 = 81% plan to attend college, so we are tempted to try and refute the 79% established by previous studies.
As the sample consists of “yes-no” answers, it can be modeled with a binomial distribution.
Now let's establish the hypothesis.
[tex]H_0[/tex]: The probability that a school senior from a certain city plan attends college after graduation is 0.79 (79%)
[tex]H_a[/tex]: The probability that a school senior from a certain city plan attends college after graduation is greater than 0.79
The binomial distribution we are going to use is the model for [tex]H_0[/tex]
[tex]P(k,200)=\binom{200}{k}0.79^k0.21^{200-k}[/tex]
where P(k,200) is the probability of getting exactly k “yes” in 200 interviews.
Since the level of significance is 5% and in the sample we got 162 “yes”, we want to show that
S = P(1,200) + P(2,200)+P(3,300)+...+P(162,200) is greater than 0.95.
If it is, then we can refute the null hypothesis and accept 0.81 as the new probability.
We can use either a cumulative binomial distribution table or the computer and we find that S=0.7807.
Since S<0.95 we cannot refute [tex]H_0[/tex]
It is worth noticing the the critical value here is 167. That is to say, if we had obtained 167 “yes” instead of 162, we could have rejected the null.
