Question

Lisa, a dentist, believes not enough teenagers floss daily. She would like to test the claim that the proportion of teenagers who floss twice a day is less than 40%. To test this claim, a group of 400 teenagers are randomly selected and its determined that 149 floss twice a day.

Answer 1

Answer:

$z=\frac{0.3725 -0.4}{\sqrt{\frac{0.4(1-0.4)}{400}}}=-1.123$  

The p value for a left tailed test would be:  

$p_v =P(z$  

Since the p value is very higher we can conclude that the true proportion  of teenagers who floss twice a day is NOT less than 40%.

Step-by-step explanation:

Information given

n=400 represent the random sample given

X=149 represent the floss twice a day

$\hat p=\frac{149}{400}=0.3725$ estimated proportion of floss twice a day

$p_o=0.4$ is the value the proportion that we want to check

z would represent the statistic

$p_v$ represent the p value

System of hypothesis

We want to check proportion of teenagers who floss twice a day is less than 40%, so then the system of hypothesis are.:  

Null hypothesis:$p \geq 0.4$  

Alternative hypothesis:$p < 0.4$  

For the one sample proportion test the statistic is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)  

If we replace the info given we got:

$z=\frac{0.3725 -0.4}{\sqrt{\frac{0.4(1-0.4)}{400}}}=-1.123$  

The p value for a left tailed test would be:  

$p_v =P(z$  

Since the p value is very higher we can conclude that the true proportion  of teenagers who floss twice a day is NOT less than 40%.