Question

A statistics instructor believes that fewer than 20% of Evergreen Valley College (EVC) students attended the opening night midnight showing of the latest Harry Potter movie. She surveys 84 of her students and finds that 11 of them attended the midnight showing. At a 1% level of significance, an appropriate conclusion is:

Answer 1

20% of Evergreen Valley college students attended the opening night midnight showing of the latest harry potter movie.

Answer 2

Answer: It is believed that exactly 20% of Evergreen Valley college students attended the opening night midnight showing of the latest harry potter movie.

Step-by-step explanation:

Since we have given that

n = 84

x = 11

So, $\hat{p}=\dfrac{x}{n}=\dfrac{11}{84}=0.13$

p = 0.20

So, hypothesis:

$H_0:p=\hat{p}\\\\H_a:\hat{p}$

so, test statistic value would be

$z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}\\\\z=\dfrac{0.13-0.20}{\sqrt{\dfrac{0.2\times 0.8}{84}}}\\\\z=-1.604$

At 1% level of significance, critical value would be

z= 2.58

Since 2.58>-1.604

So, We will accept the null hypothesis.

Hence, It is believed that exactly 20% of Evergreen Valley college students attended the opening night midnight showing of the latest harry potter movie.