Download Testing Hypothesis for mean and proportion : Homework 10 Testing

Testing Hypothesis for mean and proportion :
Homework 10
Testing Hypothesis for mean:
7.2.5:
In a sample of 49 adolescents who served as the subjects in an immunologic study ,one
variable of interest was the diameter of skin test reaction to an antigen .The sample
mean and standard deviation were 21 and 11 respectively .Can it be concluded from
these that the population mean is less than 30? Let α= 0.05
Solution :
1- Data: α= 0.05, n = 49, X =21 ,S = 11 ,  0 = 30
2- Hypothesis: H0: μ = 30
HA: μ < 30
3- Assumption : not normal , variance unknown
4- Test Statistic :
Z
X - o
21  30

 5.727
S
11 / 49
n
5- Decision Rule: Reject H0 if Z< - Z 1-α
Where Z 1-α= Z0.95=1.645
Since
Z= -5.727 < - Z 1-α= -1.645
6- Conclusion : we reject H0
mean is less than 30
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
mean is not less than 2000
7.2.17:
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
7.2.18:
A research team is willing assume that systolic blood pressure in a certain population of males
are approximately normally distributed with standard deviation of 16 . A simple random sample
of 64 males from the population had a mean systolic blood pressure reading of 133 .At the 0.05
Level of significant(α= 0.05) ,do we conclude that the population mean is greater than 130?
Solution :
5- Data: α= 0.05, n = 64 , X =133 ,σ = 16 ,  0 = 130
6- Hypothesis: H0: μ = 130
HA: μ > 130
7- Assumption : normal , variance known
8- Test Statistic :
Z
X - o


133  130
16 / 64
 1.5
n
5- Decision Rule: Reject H0 if Z > Z 1-α
Where Z 1-α= Z1-0.05 = Z0.95 = 1.645
Since
Z= 1.5
> Z 1-α = 1.645 (not satisfied)
2- Conclusion : we accept H0
mean is not greater than 130