The breaking strength of a fiber is required to be at least 150 psi. Past experience has indicated that the standard deviation of breaking strength is σ = 3 psi. A random sample of four specimens is tested. The

results are y1=145, y2=153, y3=150 and y4=147.

(a) State the hypotheses that you think should be tested in this experiment

(b) Test these hypotheses using α

= 0.05. What are your conclusions?

(c) Find the

P-value for the test in part (b).

(d) Construct a 95 percent confidence interval on the mean breaking strength.

Question

Mathematics

Devon

9 December, 20:07

The breaking strength of a fiber is required to be at least 150 psi. Past experience has indicated that the standard deviation of breaking strength is σ = 3 psi. A random sample of four specimens is tested. The

results are y1=145, y2=153, y3=150 and y4=147.

(a) State the hypotheses that you think should be tested in this experiment

(b) Test these hypotheses using α

= 0.05. What are your conclusions?

(c) Find the

P-value for the test in part (b).

(d) Construct a 95 percent confidence interval on the mean breaking strength.

+1

Answers (1)

Know the Answer?

Hadassah Alexander · Answer 1

(a) Null hypothesis: The breaking strength of a fiber is required to be 150 psi

Alternate hypothesis: The breaking strength of a fiber is required to be 150 psi or greater than 150 psi

(b) Fail to reject the null hypothesis. The breaking strength of a fiber is required to be 150 psi

(c) P-value is 0.025

(d) 95% confidence interval for the mean breaking strength is between 145.227 psi and 154.773 psi

Step-by-step explanation:

(a) The null hypothesis is a statement from the population parameter which is subject to testing. The alternate hypothesis is also a statement from the population parameter that negates the null hypothesis.

(b) Z = (sample mean - population mean) : sd/√n

sample mean = 145+153+150+147/4 = 148.75 psi, population mean = 150 psi, sd = 3 psi, n = 4

Z = (148.75 - 150) : 3/√4 = - 1.25 : 1.5 = - 0.833

This test is a two tailed test. The critical value at 0.05 significance level is 1.96. The region of no rejection of the null hypothesis lies between - 1.96 and 1.96 since it is a two tailed test.

Conclusion: since - 0.833 falls within the region bounded by - 1.96 and 1.96, fail to reject the null hypothesis. The breaking strength of a fiber is required to be 150 psi

(c) The cumulative area for the critical value Z = 1.96 is 0.9750

P-value = 1 - 0.9750 = 0.025

(d) Confidence Interval = mean + or - error margin (E)

E = t*sd/√n

n = 4, degree of freedom = n-1 = 4-1 = 3, t-value corresponding to 3 degrees of freedom and 95% confidence level is 3.182

E = 3.182*3/√4 = 4.773

Lower limit = mean - E = 150 - 4.773 = 145.227

Upper limit = mean + E = 150 + 4.773 = 154.773

95% confidence interval is (145.227 psi, 154.773 psi)