α = 0.05 vs p-value?

#1
So I've got a couple questions on my stats homework as follows:

E. Canis infection is a tick-bone disease of dogs that is sometimes contracted by humans. Among infected humans, the distribution of WBC counts has an unknown mean u and a SD o. In the general population, the mean WBC count is 7250/mm3. It is believed that persons infected with E. Canis must on average have lower WBC counts.

a) What are the null and alternative hypotheses for a one-sided test?

b) for a sample of 15 infected persons, the mean white blood cell count is
X(bar) = 4767/mm3 and the standard deviation is s=3204/mm3. Conduct the test at the α = 0.05 level (This is a one-sided test)

d) calculate a p-value

e) repeat parts b, c and d using two-sided testing and comment on the differences in the results that you get.


a) H0: u < (or equal to) 7250/mm3
HA: u > 7250/mm3

I think I understand that I have to calculate a one-sided test for b & c the first time around, then a 2-sided test for b & c. Here are my calculations for the test statistic for the one sided test

b) t = [X(bar) - u] / [s/(sqrt)n]
t = [4767 - 7250] / 3204 / (sqrt)15
t = -2483 / 827.2692
t = -3.001

Then, determine the critical regions:

-tn-1, 0.05
-t15-1.05
-t = -(infinity) to -2.145

therefore, we can reject the null because the value of -3.001 is within the critical region?

OR, should I be doing something like a 95% confidence interval, like I've done below?


X(bar) +/- t n-1, 0.025 x (standard error)

95% CI = 4767 +/- 2.145 x 2304/(sqrt)15
95% CI = 4767 +/- 1276.0395
95% CI = (6043.0395 , 3490.9604)

I hope that this is clear enough
 
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trinker

ggplot2orBust
#2
I don't have time to throw up an answer just a quick comment; your [MATH]H_A[/MATH], to my knowledge shouldn't have an equal sign in it.
 

lorb

New Member
#3
a) H0: u < (or equal to) 7250/mm3
HA: u > (or equal to) 7250/mm3
H0 and HA need to be disjoint. You should get rid of one of the (or equal to) clauses.


Edit: I think you t-test is fine but the CI is wrong. If you do a one sided test your CI also has just one limit and a different t-value. If you don't have tables for a one sided test (tailed) you just double alpha. So here we would have 1.765.
From that you calculate an upper/lower limit and the other limit is plus/minus infinity.

Edit2: I like this website because it has a nice visualisation that (at least for me) helps to wrap your mind about that stuff: http://www.statdistributions.com/t/?p=0.1&df=14
 
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#5
In the general population, the mean WBC count is 7250/mm3. It is believed that persons infected with E. Canis must on average have lower WBC counts.

a) What are the null and alternative hypotheses for a one-sided test?

...
a) H0: u < (or equal to) 7250/mm3
HA: u > 7250/mm3


I could be wrong here, but I think generally the experimental hypothesis is the alternative hypothesis,
not the null hypothesis. so your null hypothesis here for the one-sided test would be that
the wbc for infected individuals was equal to or higher to the wbc for the general population,
and the alternative hypothesis would be that it is lower, i.e.,

a) H0: u > (or equal to) 7250/mm3
HA: u < 7250/mm3

Also, confidence intervals correspond to two-sided p-values, so for the part of the problem
about one-sided p-values, you should just give a p-value and not a confidence interval.

You have correctly calculated the t-statistic for the one-sided test, so it is correct that you should
reject the null hypothesis. For part d, you now need to determine the associated p-value
for this t-statistic.
 
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