A rivet is to be inserted into a hole. If the standard deviation of hole diamete
ID: 3132878 • Letter: A
Question
A rivet is to be inserted into a hole. If the standard deviation of hole diameter exceeds 0.02 mm, there is an unacceptably high probability that the rivet will not fit. A random sample of n = 15 parts is selected, and the hole diameter is measured. The sample standard deviation of the hole diameter measurements is s = 0.016 mm. Is there strong evidence to indicate that the standard deviation of hole diameter exceeds 0.02 mm? Calculate (a) lower bound and upper bound of P-value to draw conclusions. Round your answers to 1 decimal. Answers are exact. Construct a 95% lower confidence bound for sigma. Round your answer to 3 decimal places. Use the confidence bound in part (a) to test the hypothesis (if reject enter a value of 1, if not then enter a value of 2).Explanation / Answer
a. Assuming the population to be normally distributed, to test a hypothesis about the value of the standard deviation of a population, the test statistic will follow a chi-square distribution with (n-1) degrees of freedom, and is given by the formula:
2 = (n-1) * s2 / 02
Where n is the sample size, s is the sample standard deviation and 0 is the hypothesized value.
As per the problem, the value of the test statistic is:
214 = (15 - 1) * 0.0162 / 0.022
= 14 * 0.000256 / 0.0004
= 8.96
We need to check the 2 table to find between what p-values this 2 value lies (against a d.f. of 14). As verified from the 2 table, 8.96 lies between the 2 values for p = 0.90 (2 = 7.790) and p = 0.75 (2 = 10.165).
Therefore, the lower and upper bounds of p-value to draw conclusions are 0.9 and 0.75 0.8 respectively.
Please note, as we are interested in the case that the S.D. of the diameter exceeds 0.02, one-sided p-values have been considered.
b. A (1 – ) level confidence interval for is given by [ { (n-1)s2 / 2/2 , n-1 }1/2 , { (n-1)s2 / 21- /2 , n-1 }1/2 ]
Therefore, a 95% lower confidence bound for
= { (n-1)s2 / 21- /2 , n-1 }1/2
= { (15-1) * 0.0162 / 20.975 , 14 }1/2 Since, = 1 – 0.95 = 0.05
= { 14 * 0.000256 / 26.119 }1/2
= 0.012
c. Suppose, we want to test the hypothesis H0: 0.02 (against H1: > 0.02) at a 5% level of significance.
Since, we are considering a right-handed test, the p-value is 0.05. Since the upper bound of p-value calculated in part (a) is 0.75 > 0.05, we cannot reject the null hypothesis at a 0.05 level of significance.
In other words, based on the sample evidence, we cannot conclude that the standard deviation of the hole diameter exceeds 0.02 mm.
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