Question

The Director of Student Activities is concerned that she has not hired enough security for the Sadie Hawkins Day dance, and wants to know approximately how many students will be attending. If more than 25% of the student body is planning on going to the dance, she will need to hire more security. She sends out a survey to 300 students, and out of the 285 who respond, 82 say they plan to attend.

On the basis of this survey, should the Director of Student Activities hire more security? Use appropriate statistical evidence to support your conclusion.

There were 15 people who didn't return their surveys. Suppose they had.

Is it possible that the Student Activities Director's decision would have been different?

Answer 1

Assuming 0.05 significance level, there is no significant evidence, on the basis of this survey, that the Director of Student Activities has to hire more security.

If 15 people who didn't return their surveys had returned their surveys and all reported attending the dance, Student Activities Director's decision would have been different, because there would be significant evidence on the basis of this survey, that the Director of Student Activities has to hire more security.

Explanation:

Lep p be the proportion of students who will attend the Sadie Hawkins Day dance.

Null and alternative hypotheses are:

• 
  `H_(0):` p=0.25

• 
  `H_(a):` p>0.25

Test statistic can be calculated using the equation:

z=
`\frac{p(s)-p}{\sqrt{(p*(1-p))/(N) } }` where

• p(s) is the sample proportion of students who will attend the Sadie Hawkins Day dance (
  `(82)/(285)` ≈0.288)

• p is the proportion assumed under null hypothesis. (0.25)

• N is the sample size (285)

Then z=
`\frac{0.288-0.25}{\sqrt{(0.25*0.75)/(285) } }` ≈ 1.48

Corresponding one tailed p-value for the test statistics is ≈ 0.069

Assuming a significance level 0.05, the result is not significant since 0.069>0.05.

Therefore there is not significant statistical evidence at 0.05 significance level that the director has to hire more security.

Suppose 15 people who didn't return their surveys had returned their surveys and all reported attending the dance.

Then above sample proportion of students who will attend the Sadie Hawkins Day dance changed as (
`(97)/(300)` ≈0.323). Sample size would be 300.

If we recalculate the test statistic:

z=
`\frac{0.323-0.25}{\sqrt{(0.25*0.75)/(300) } }` ≈ 2.92

And p-value is ≈ 0.00175. This result is significant at 0.05 significance level,

Therefore in this case (if 15 people who didn't return their surveys had returned their surveys and all reported attending the dance) Student Activities Director's decision would have been different