One method for straightening wire prior to coiling it to make a spring is called

Question

One method for straightening wire prior to coiling it to make a spring is called "roller straightening''. Suppose that a sample of 16 wires is selected and each is tested to determine tensile strength (N/mm^2). The resulting sample mean and sample standard deviation are 2164.4 and 30.5, respectively. It is known that the mean tensile strength for spring made using spinner straightening is 2150 N/mm^2. (1) What is the random variable X in this problem? What does the mean mu of X represent? (2) What null hypothesis and alternative hypothesis should be tested in order to determine if the mean tensile strength for the roller method is better than the mean tensile strength for spinner method? (3) Is this one-tailed or two-tailed test? (4) What test statistic should be used to test the hypotheses? Is a normality assumption of the population necessary? Why? (5) At the significance level alpha = 0.05, compute the rejection region (RR). (6) Compute the value of your test statistic (assuming H_0). (7) What is the conclusion of your test? Explain in your own words, avoiding statistical terms as possible.

Explanation / Answer

1) The random variable X in this problem is tensile strength measured in N/mm2. Here sample mean is 2164.4 and sample standard deviation is 30.5. Here the sample mean needs to be compared with the mean tensile strength for spinner method.

2) Null hypothesis = The mean tensile strength for the roller method is same as the mean tensile strength for spinner method.

Alternative hypothesis = The mean tensile strength for the roller method is better than the mean tensile strength for spinner method.

3) This is a one tailed test as we are interesting in determining whether the mean tesnile strength of roller method is greater than the mean tensile strength for spnner method.

4) As the sample mean is small, less than 30, t-statistic should be used to test the hypothesis. Yes normality assumption of the population is important because thick tailed or heavily skewed distributions can considerably reduce the power of the test.

5) and 6) Significane level = 0.05

t statistic = (observed - expected)/SE

= (2164.4 - 2150)/(30.5/sqrt(16))

= 14.4/7.625 = 1.888525

Degrees of freedom = 16 - 1 = 15

From t table we can calculate the critical value corresponding to 15 degrees of freedom at 0.05 significance level,

critical value = 1.753

So critical region is region to the right of 1.753 in the t distribution with 15 degrees of freedom.

7) As the t statistic 1.888525 is greater than  1.753, we can reject the null hypothesis. That means the  mean tensile strength for the roller method is better than the mean tensile strength for spinner method.

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.