Question

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?

Answer 1

Answer:

(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)