Question

A cutting machine is adjusted so that the mean of a certain dimension of the parts is 20 cm. A random sample of 33 dimensions revealed a sample mean of 20.33 cm with a variance of 0.04 cm. Do these results indicate that the machine is out of adjustment?

Answer 1

Let $\mu$ denotes the mean of a certain dimension of the parts.

To test null hypothesis Ho :u20.0 against alternative hypothesis H1u20.0 Let denotes the sample mean, s denotes the sample standard deviation and n denotes the sample size. Here x = 20.3300, s = 0.0400, s = 0.2000, n = 33 The test statistic can be written as Vn (20.0) t= S which under Ho follows a t distribution with n -1 df. We reject Ho at 0.050 level of significance if P-value 0.050 or if|tobs| > to.025,n-1 Now, The value of the test statistic 9.478528 associated degrees of freedom = 32 The critical value = 2.036933 P- value P(|t,_-1| > tobs) = 2* P(t32 > 9.478528) = 2* (1 - 1.000000) = 0.000000 Since p-value< 0.050 and tobs >tcritical = 2.036933, so we reject Ho at 0.050 level of significance Hence, we can conclude that the the population mean is not equal to 20.0.