Are the results statistically significant as claimed Use the

Are the results statistically significant as claimed? Use the \"No Adolescent Marijuana\" group to determine a suitable p_0 and then test the hypothesis: H_0 :p p_0

Solution

By using the variable no adolescent Marijuana use:

hypothesis is,

H0 : p p0 Vs H1 : p > p0

The test is one sided and right tailed.

alpha = level of significance = 0.05

From the graph we can see that percent developing schizophrenia is 2.3% for no adolescent Marijuana use.

2.3% is obtain from data so 2.3% is the sample proportion (p^).

n = 311

Now we have to find population proportion.

The test is right sided so we can take value of population proportion is any value which is less than 2.3%.

p^ = 2.3% = 2.3/100 = 0.023

p = 0.011

q = 1 - p = 1 - 0.011 = 0.989

The test statistic for testing single proportion is,

Z = (p^ - p) / sqrt((p*q)/n)

Z = (0.023 - 0.011) / sqrt((0.011*0.989)/311)

Z = 2.0289

We can take decision by using P-value.

EXCEL sytax for P-value is,

=1 - NORMSDIST(z) (the test is right tailed)

where, z is the test statistic value.

P-value = 0.02

P-value < alpha

Reject H0 at 5% level of significance.

The results are statistically significant.