A recent drug survey showed an increase in use of drugs and alcohol among local high school students as compared to the national percent. Suppose that a survey of 100 local youths and 100 national youths is conducted to see if the percentage of drug and alcohol use is higher locally than nationally. Locally, 65 seniors reported using drugs or alcohol within the past month, while 61 national seniors reported using them. Required:
Conduct a hypothesis test at the 5% level.

We fail to reject H0

Step-by-step explanation:

Null hypothesis : H0 : Plocal ≤ Pnational

Alternative hypothesis : H1 : Plocal > Pnational

Plocal = P1 = 65/100 = 0.65

Pnational = P2 = 61/100 = 0.61

n1 = 100 ; n2 = 100

Mean, m = P1 - P2 = 0.65 - 0.61 = 0.04

The standard deviation : = sqrt(Variance)

Variance = P1(1 - P1) /n1 + P2(1 - P2) /n2

Variance = 0.65(0.35)/100 + 0.61(0.39)/100

Variance = 0.002275 + 0.002379 = 0.004654

Standard deviation (s) = √(0.004654) = 0.06822

Z = m/s = 0.04 / 0.06822 = 0.586

Using the p-value from Z score calculator :

Hence, Pvalue for a Zscore 0.586 at 5% = 0.278938

Since ;

P-value is > α ; 0.278938 > 0.05

Then we fail to reject H0

c

step-by-step explanation: