Answer:
a.H0 : u1= u2 against Ha : u1≠ u2 This is a two sided test
b) There isn't enough evidence that the machine is not meeting the manufacturer’s specifications in terms of the average breaking strength.
c) the p- value is 0.0359*2= 0.0718. It is greater than the value of ∝ so there isn't enough evidence that the machine is not meeting the manufacturer’s specifications in terms of the average breaking strength.
d) There is enough evidence that the machine is not meeting the manufacturer’s specifications in terms of the average breaking strength.
e) There isn't enough evidence that the machine is not meeting the manufacturer’s specifications in terms of the average breaking strength.
Step-by-step explanation:
Formulate the null and alternative hypotheses as
a) H0 : u1= u2 against Ha : u1≠ u2 This is a two sided test
Here ∝= 0.005
For alpha by 2 for a two tailed test Z∝/2 = ± 1.96
Standard deviation = s= 3.5 pounds
n= 49
The test statistic used here is
Z = x- x`/ s/√n
Z= 69.1- 70 / 3.5 / √49
Z= -1.80
Since the calculated value of Z= -1.80 falls in the critical region we reject the null hypothesis.
b) There isn't enough evidence that the machine is not meeting the manufacturer’s specifications in terms of the average breaking strength.
c) the p- value is 0.0359*2= 0.0718. It is greater than the value of ∝ so there isn't enough evidence that the machine is not meeting the manufacturer’s specifications in terms of the average breaking strength.
d) If standard deviation is 1.75 pounds
The test statistic used here is
Z = x- x`/ s/√n
Z= 69.1- 70 / 1.75 / √49
Z= -3.6
This value does not fall in the critical region.
d) There is enough evidence that the machine is not meeting the manufacturer’s specifications in terms of the average breaking strength.
e) If the sample mean is 69 pounds
Z = x- x`/ s/√n
Z= 69.1- 69 / 3.5 / √49
Z= 0.2
This value falls in the critical region
e) There isn't enough evidence that the machine is not meeting the manufacturer’s specifications in terms of the average breaking strength.
