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Classroom Example for Tests of Hypotheses Involving a Proportion.
The problem was to test the statement that our salesperson makes a sale on at least 50% of calls. To test this
with a significance level of 5% we send the salesperson out and find that he/she makes only 40 sales in 100
calls. The excerpt from the formula table on tests of proportions is stated below.
Interval for
Proportion
Confidence
Interval
p  p  z 2 s p
pq
n
q  1 p
sp 
Hypotheses
Test Ratio
H 0 : p  p0
z
H1 : p  p0
p  p0
p
Critical Value
p cv  p0  z 2 p
p 
p0 q0
n
Our hypotheses are H 0 : p  .50 H 1 : p  .50 .   .05, n  100 , p 0  .5 , q 0  1  p 0  1  .5  .5.
p  p0
a) We use the test ratio z 
. To do this we need the standard deviation of the sample proportion
p
p0 q0
.5.5

 0.05 . Because the alternate hypothesis says p  .50 , we are worried about
n
100
p  p0
p  .5

negative. We make a diagram of the Normal
p  .50 , which would make our test ratio z 
p
.05
p 
distribution with zero in the middle showing that our rejection region is below z .05  1.645 .
We now use the fact that there were x  40 successes to compute p 
x 40

 .40 , so that
n 100
.40  .5
 2 . Since this value of z is in our rejection region, we reject the null hypothesis.
.05
Alternately we find a p-value for z . In the case of an alternate hypothesis like the one above, the p-value is
Px  40  Pp  .40  Pz  2.00  .5  .4772  .0228 . The rule on p-value(from the website) says if
the p-value is less than the significance level (alpha) reject the null hypothesis. Since .0228 is below
.05, we reject the null hypothesis.
b)We use the critical value for the proportion. Since the alternative hypothesis is H 1 : p  .50 , we are

worried about the proportion being too low, so we want a critical value for p that is below .50. The critical
z
value formula, pcv  p0  z  p must be modified to give us a number below .50, so. Make a diagram
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with the mean at .50. pcv  p0  z  p  .50 1.645.05  .4178. Show that the rejection zone is the area
below .4178. Since p  .40 is below this critical value, reject H 0 .
c) To do a confidence interval, modify p  p  z s p where q  1  p  1  .40  .60 and
2
pq
.40 .60 

 .0024  .0490 in the direction of the alternate hypothesis H 1 : p  .50 to say
n
100
p  p  z s p  .40  1.645.0490 or p  .4806 . Since this contradicts H 0 : p  .50 , reject H 0 .
sp 
© 2002 R. E. Bove
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