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6 July, 02:31

The mean breaking strength of a fiber is required to be at least 150 psi. Past experience has indicated that the standard deviation of the breaking strength is σ = 3 psi. A random sample of four specimen are tested and the results are:X_1=145, X_2=153, X_3=150 and X_4=142. a. State the hypothesis to be tested?

b. Test the hypothesis and state your conclusions? Use α=0.05?

c. Should the P value be less than α in this case? Explain.

d. Construct the 95% confidence interval on the true mean breaking strength?

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  1. 6 July, 02:56
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    (a) Null hypothesis: The mean breaking strength of a fiber is 150 psi.

    Alternate hypothesis: The mean breaking strength of a fiber is less than or equal to 150 psi.

    (b) The test statistic is - 1.67. There is sufficient evidence to conclude that the mean breaking strength of a fiber is 150 psi.

    (c) No, the P value should not be less than the significance level (alpha).

    (d) 95% confidence interval for the true mean breaking strength is between a lower limit of 140.71 psi and an upper limit of 154.29 psi.

    Step-by-step explanation:

    (a) A null hypothesis is a statement from a population parameter which is either rejected or accepted (fail to reject) upon testing. It is expressed using the equality sign.

    An alternate hypothesis is also a statement from the population parameter negates the null hypothesis and is accepted if the null hypothesis is rejected. It is expressed using any of the inequality signs.

    (b) sample mean = (145+153+150+142) / 4 = 590/4 = 147.5 psi

    population mean = 150 psi

    population sd (sigma) = 3 psia

    n = 4

    Test statistic (z) = (sample mean - population mean) : (population sd/√n) = (147.5 - 150) : (3/√4) = - 2.5 : 1.5 = - 1.67

    The test is a two-tailed test because the alternate hypothesis is expressed using less than or equal to. For a 5% significance level, critical values are - 1.96 and 1.96.

    Conclusion:

    Fail to reject the null hypothesis because the test statistic - 1.67 falls within the region bounded by the critical values.

    (c) The P value should not be less than alpha because to fail to reject the null hypothesis using the P value approach, the P value must be greater than alpha (significance level).

    (d) Confidence interval is given as mean + / - margin of error (E)

    mean = 147.5 psi

    sample sd = sqrt[ ((145-147.5) ^2 + (153-147.5) ^2 + (150-147.5) ^2 + (142-147.5) ^2) : 4] = 4.27 psi

    n = 4

    degree of freedom = n - 1 = 4-1 = 3

    confidence level = 95%

    critical value (t) corresponding to 3 degrees of freedom and 95% confidence level is 3.182

    E = t * sample sd/√n = 3.182*4.27/√4 = 6.79 psi

    Lower limit = mean - E = 147.5 - 6.79 = 140.71 psi

    Upper limit = mean + E = 147.5 + 6.79 = 154.29 psi

    95% confidence interval is (140.71, 154.29)
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