Q.1.1 Check your understanding - The Practice of Statistics for AP

Question

According to the National Campaign to Prevent Teen and Unplanned Pregnancy, 20% of teens aged 13 to 19 say that they have electronically sent or posted sexually suggestive images of themselves. 'The counsellor at a large high school worries that the actual figure might be higher at her school. To find out, she gives an anonymous survey to a random sample of 250 of the school's 2800 students. All 250 respond and 63 admit to sending or posting sexual images. Carry out a significance test at the $\alpha =0.05$ significance level. What conclusion should the counsellor draw?

Step-by-Step Solution

Answer

The conclusion is that the actual figure is higher than assumed before

Step 1: Introduction

Teen pregnancy is the point at which a lady under twenty gets pregnant. It ordinarily alludes to adolescents between the ages of $15-19$. However, it can incorporate young ladies as youthful as ten. It's likewise called juvenile pregnancy.

Step 2: Explanation

Given,

Sample size n is $250$ and Confidence level is $95\%$

Significance level $\alpha =0.05$ and population proportion is $0.20$

Calculating the null and alternative hypotheses,

$\begin{array}{r}{H}_0:p=0.20\\ H_a:p>0.20\end{array}$

Using,

$$\begin{gathered} z=\frac{\widehat{p}-p_0}{\sqrt{\frac{p_0\left(1-p_0\right)}{n}}} \\ =\frac{0.252-0.019}{\sqrt{{\displaystyle \frac{0.019(1-0.019)}{250}}}}=2.055 \end{gathered}$$

Hence the conclusion is that the actual figure is higher than assumed before