Question

According to a survey by the CDC, about 18% of high school students in the U.S. smoked in 2009, down from 22% in 2003. Suppose in a recent random sample of 600 high school students, 90 said that they smoked. At the 5% level of significance, can you conclude that the current percentage of U.S. high school students who smoke is different from 18%? Use the p-value test. (Fully explain why, of course, ‘mathematically’.)

Answer #1

Answer)

Null hypothesis Ho : P = 18%

Alternate hypothesis Ha : P not equal to 18%

N = 600

P = 0.18

First we need to check the conditions of normality that is if n*p and n*(1-p) both are greater than 5 or not

N*p = 108

N*(1-p) = 492

Both the conditions are met so we can use standard normal z table to estimate the P-Value

Test statistics z = (oberved p - claimed p)/standard error

Standard error = √{claimed p*(1-claimed p)/√n

Observed P = 90/600

Claimed P = 0.18

N = 600

After substitution

Test statistics z = -1.91

From z table, P(z<-1.91) = 0.0281

But our test is two tailed

So, P-Value is = 2*0.0281 = 0.0562

As the obtained P-Value is greater than 0.05 (given significance level)

We fail to reject the null hypothesis Ho

So we do not have enough evidence to conclude that proportion of high school students who smoke is different from 18%

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