Suppose that the director of manufacturing at a clothing factory needs to determine whether a new machine is producing a particular type of cloth according to the manufacturer’s specifications, which indicate that the cloth should have a mean breaking strength of 70 pounds and a standard deviation of 3.5 pounds. A sample of 49 pieces reveals a sample mean of 69.1 pounds.

(a) State the null and alternative hypotheses.

(b) Is there evidence that the machine is not meeting the manufacturer’s specifications in terms of the average breaking strength? (Use a 0.05 level of significance.)

(c) Compute the p-value and interpret its meaning.

(d) What will your answer be in (b) if the standard deviation is 1.75 pounds?

(e) What will your answer be in (b) if the sample mean is 69 pounds?

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.