7.2.8:
We wish to know if we can conclude that the mean daily caloric intake in the adult
rural population of a developing country is less than 2000. A sample of 500 had a
mean of 1985 and a standard deviation of 210 . Let α= 0.05
Solution :
1- Data: α= 0.05, n = 500, X =1985 ,S = 210 ,  0 = 2000
2-Hypothesis: H0: μ = 2000
HA: μ < 2000
3-Assumption : not normal , variance unknown
4-Test Statistic :
Z
X -  o 1985  2000

 1.597
S
210 / 500
n
5- Decision Rule: Reject H0 if Z< - Z 1-α
Where Z 1-α= Z0.95=1.645
Since
Z= -1.597 < - Z 1-α= -1.645
6-Conclusion : we accept H0
7.2.17:
mean is not less than 2000
Suppose it is known that the IQ scores of a certain population of adults are
approximately normally distributed with standard deviation of 15 .A simple of
random sample of 25 adults drawn from this population had mean 105 . On the
basis of these data can we conclude that the mean IQ score is not 100. Let the
probability of type I error be 0.05 (α= 0.05)
Solution :
1-Data: α= 0.05, n = 25 , X =105 ,σ = 15 ,  0 = 100
2-Hypothesis: H0: μ = 100
HA: μ ≠ 100
3-Assumption : normal , variance known
4-Test Statistic :
Z
X - o


105  100
15 / 25
 1.67
n
5- Decision Rule: Reject H0 if Z< - Z 1-α/2 or Z > Z 1-α/2
Where Z 1-α/2= Z1-0.05/2 = Z0.975 = 1.96
Since
Z= 1.67 > Z 1-α/2 = 1.96 or Z=1.67 < - Z 1-α/2 = -1.96 (not satisfied)
6-Conclusion : we accept H0
mean is equal than 100
7.5.1:
Jacquemyn conducted a survey among gynecologists-obstetrician in the Flanders region
and obtained 295 responses . Of those responding ,90 indicated that they had performed at
least one caesarean section on demand every year. Does this study provide sufficient
evident for us to conclude that less than 35 percent of the gynecologists-obstetrician in the
Flanders region caesarean section on demand every year? α= 0.05
Solution :
a
90
1-Data: α= 0.05, n = 295 , a = 90 , Pˆ  
 0.31 , P0 = 35/100=0.35
n 295
2-Hypothesis: H0: P = 0.35
HA: P < 0.35
3-Test Statistic :
Z
Pˆ  P0
p0 q0
n

0.31  0.35
0.35(0.65)
295
 1.433
4- Decision Rule: Reject H0 if Z < - Z 1-α
Where Z 1-α= Z1-0.05 = Z0.95 = 1.645
Since
Z= -1.433
< - Z 1-α = -1.645 (not satisfied)
5-Conclusion : we accept H0
proportion is not smaller than 0.35