Question

Begnning in 1988 through 2013, the yearly numbers of publications in referred journals by Professors Swartz are as follows: 2,0,0,1,5,1,2,4,2,3,4,2,3,3,2,2,4,2,2,5,1,3,2,6,1,3. In a test of whether his publication rate is 3 papers per year, the p-value = 2P(X<= 2.5) = 2 P( t26 <= (2.5-3)/(1.5033/ sqrt (26) ) = 2P ( t25 <= -1.696) = 0.102 is obtained and the null hypothesis is not rejected . There are various probelms with the assumptions underlying the test procedure. Describe some of the problems.

Answer 1

Sol:

H0:mu=3

H1:mu not =3

alpha=0.05

Rcode to get t statistic and p value is:

no_of_publications <- c(2,0,0,1,5,1,2,4,2,3,4,2,3,3,2,2,4,2,2,5,1,3,2,6,1,3)
t.test(no_of_publications,mu=3)

Result:

One Sample t-test

data: no_of_publications
t = -1.6959, df = 25, p-value = 0.1023
alternative hypothesis: true mean is not equal to 3
95 percent confidence interval:
1.892792 3.107208
sample estimates:
mean of x
2.5

t=-1.6959

p=0.1023

p>0.05

do not reject H0

Accept Null hypothesis.

various probelms with the assumptions underlying the test procedure:

NORMALITY ASSUMPTION:

If the distribution of sample data of yearly number of publicatons is normal ,fine to use a t test.

But if the underlying distribution is not normal and sample size is small(rule of thumb n>30 per group if not too skewed;n>100 if the distribution is really skewed,the central imit theorem takes some time to kick in.cannot use t test

t test is very robust against the normality assumption.

if its not normal use wilcoxson rank sum test

INDEPENDENCE (CRITICAL)

For single mean as in this case,the observations within a sample must be indpendent.