The number of successes and the sample size for a simple random sample from a population are given below.
xequals=3535,
nequals=5050,
Upper H 0H0:
pequals=0.60.6,
Upper H Subscript aHa:
pgreater than>0.60.6,
alphaαequals=0.050.05
a. Determine the sample proportion.
b. Decide whether using the one-proportion z-test is appropriate.
c. If appropriate, use the one-proportion z-test to perform the specified hypothesis test.
Solution :
This is the right tailed test .
The null and alternative hypothesis is
H0 : p = 0.60
Ha : p > 0.60
n = 50
x = 35
a )
= x / n = 35 / 50 =0.70
P0 = 0.60
1 - P0 = 1 - 0.60 = 0.40
b ) Test statistic = z
=
- P0 / [
P0
* (1 - P0 ) / n]
=0.70 - 0.60 / [0.60
*0.40 / 50 ]
= 1.44
Test statistic = z =1.44
P(z > 1.44 ) = 1 - P(z < 1.44 ) = 1 - 0.9251
c ) P-value = 0.0749
= 0.05
P-value >
0.0749 > 0.05
Do not reject the null hypothesis .
There is insufficient evidence to suggest that
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