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30 August, 14:32

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.

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Answers (1)
  1. 30 August, 15:43
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    (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)
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