Question

The breaking strength of cables produced by manufacture have a mean value of 1800 pound (lb) and a standard deviation of 100 lb. By a new technique in the manufacturing process, it is claimed that the breaking strength can be increased. To test this claim, a sample of 50 cables is tested and it is found that the mean breaking strength is 1850 lb. Based on the above information, answer the following four questions:

a) Formulate your hypothesis and write the meaning of the hypothesis

b) Perform your hypothesis test with 95% and 99% confidence interval

c) Construct a 95% and 99% confidence intervals

d) Make your final conclusion

Answer 1

SOLUTION:

From given data,

The breaking strength of cables produced by manufacture have a mean value of 1800 pound (lb) and a standard deviation of 100 lb. By a new technique in the manufacturing process, it is claimed that the breaking strength can be increased. To test this claim, a sample of 50 cables is tested and it is found that the mean breaking strength is 1850 lb. Based on the above information, answer the following four questions:

Where,

Mean = = 1800

Standard deviation = S = 100

Sample size = n = 50

Sample mean = = 1850

a) Formulate your hypothesis and write the meaning of the hypothesis

Test hypothesis:

H₀ :   = 1800 (null hypothesis)

H_a :   > 1800 (Alternative hypothesis)

The meaning of the hypothesis

Hypothesis is an assumption about the population parameter

b) Perform your hypothesis test with 95% and 99% confidence interval

Test statistics:

Z = (X-) / (S/sqrt(n))

Z = (1850-1800) / (100/sqrt(50))

Z = 3.5355

P-value:

P-value = P(Z > 3.5355)

P-value = 1 - P(Z < 3.5355)

P-value = 1 - 0.99977

P-value = 0.0002

95% confidence interval

Confidence interval is 95%

95% = 95/100 = 0.95

= 1 - Confidence interval = 1-0.95 = 0.05

P-value = 0.0002 <   = 0.05

99% confidence interval

Confidence interval is 99%

99% = 99/100 = 0.99

= 1 - Confidence interval = 1-0.99 = 0.01

P-value = 0.0002 <   = 0.01

We reject the null hypothesis at 99% and 95% .

c) Construct a 95% and 99% confidence intervals

95% confidence interval

Confidence interval is 95%

95% = 95/100 = 0.95

= 1 - Confidence interval = 1-0.95 = 0.05

/2 = 0.05 / 2

= 0.025

Z_/2 = Z_0.025 = 1.96

for the population mean µ

   Z_/2 (S / )

   Z_0.025 (S / )

1850   1.96 (100 / )

1850 1.96 (100 /7.0710678)

1850 1.96 14.1421356

1850 27.71858

(1850-27.71858 , 1850+27.71858)

(1822.28 , 1877.71)

99% confidence interval

Confidence interval is 99%

99% = 99/100 = 0.99

= 1 - Confidence interval = 1-0.99 = 0.01

/2 = 0.01 / 2

= 0.005

Z_/2 = Z_0.005  = 2.58

   Z_/2 (S / )

   Z_0.005   (S / )

1850   2.58 (100 / )

1850 2.58 (100 /7.0710678)

1850 2.58 14.1421356

1850 36.48670

(1850-36.48670 , 1850+36.48670)

(1813.51 , 1886.48)

d) Make your final conclusion

We reject the null hypothesis at 99% and 95% .