Results from previous studies showed 79% 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. Does this indicate that the percentage has increased from that of previous studies? Test at the 5% level of significance. Compute the z or t value of the sample test statistic

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.