According to a publication 13.5% of 18 to 25 year olds were users of marijuana i

Question

According to a publication 13.5% of 18 to 25 year olds were users of marijuana in 2000. A recent poll of 1398 randomly selected 18 to 25 year olds revealed that 214 currently use marijuana. At the 20% significance level, do the data provide sufficient evidence to conclude that the percentage of 18 to 25 year olds who currently use marijuana has changed from the 2000 percentage of 13.5%? Use the one proportion z-test to perform the appropriate hypothesis test, after checking the conditions for the procedure.
What are the hypotheses for the one proportion z-test?
What is the test statistic z=
Identify the critical value(s)=
Reject or do not reject?

Explanation / Answer

Here, n = 1398, p = 0.135. Hence,

np(1-p) = 1398*0.135*(1-0.135) = 163.25145 > 10, hence, we can use normal approximation.

Formulating the null and alternatuve hypotheses,          
          
Ho:   p   =   0.135
Ha:   p   =/=   0.135 [HYPOTHESES]

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As we see, the hypothesized po =   0.135      
Getting the point estimate of p, p^,          
          
p^ = x / n =    0.153075823      
          
Getting the standard error of p^, sp,          
          
sp = sqrt[po (1 - po)/n] =    0.009139477      
          
Getting the z statistic,          
          
z = (p^ - po)/sp =    1.97777425   [TEST STATISTIC]

*******************************  
          
As this is a    2   tailed test, then, getting the critical values,  
          
zcrit = -1.28, 1.28 [ANSWER, CRITICAL VALUES]      

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As |z| > 1.28, then WE REJECT THE NULL HYPOTHESIS. [ANSWER]

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