Question

QUESTION 2:

When a manufacturing process is operating properly, maximum length of a certain part must be 25cm. The lengths which are more than 25cm are scrap. If the proportion of scrap is more than 0.171255, then a modification of the process is needed to reduce this amount. On one production run, a random sample of 50 consecutive parts are selected and the following lengths are observed.

19.52	23.15	23.17	18.33	25.68	18.51	20.63	25.68	21.21	26.58
19.85	19.63	20.58	20.75	22.98	19.65	19.58	22.34	25.69	25.52
22.56	20.54	20.14	23.91	25.32	22.27	26.71	24.63	24.86	18.82
25.21	22.47	23.27	25.39	26.78	20.48	22.37	20.38	18.10	18.02
26.84	20.85	21.40	20.40	22.33	18.11	25.36	21.42	26.77	19.08

1. Does the sample indicate that process has changed and must be modified, or is it still reasonable to say that the machines are operating properly? Use a = 0.01.
   1. What is parameter of interest?
   2. Find a point estimate for the parameter?
   3. Formulate the hypotheses.
   4. Write test statistics and calculate its value.
   5. What is p value?
   6. Write rejection criteria.
   7. Write your conclusion.

2. Answer the question in part a) by critical value approach. Comment upon your result. Compare your answer with part a). Are they different? Why?

3. Calculate appropriate confidence bound to test the claim. State your result and draw your conclusion about the hypothesis.

Answer 1

Here we are given when a manufacturing process is operating properly, maximum length of a certain part must be 25cm. The lengths which are more than 25cm are scrap. If the proportion of scrap is more than 0.171255, then a modification of the process is needed to reduce this amount.

First we count f = number of scrap that is length more than 25 cm

= 13

a) i) Here, the parameter of interest is the proportion of scrap.

ii) The point estimate of the parameter = f/n = 13/50 = 0.26

iii) We consider the following hypothesis-

H₀: p = 0.171255 vs H₁: p > 0.171255

iv) The test statistic is -

  

= 1.67

v) To obtain the p-value we perform the following steps in minitab-

1. Click on “Stat” then “Basic statistic” then “1-proportion”

2. Select “summarized data” and enter 13 as “no.of events” and 50 as “no.of trials”.

3. Select “perform hypothesis test” and enter the value of given proportion in “hypothesized proportion”.

4. Click on “options” and set 99.0 as confidence level and select more than as alternative and then click OK.

5. Click OK.

Thus we, get the p-value as 0.048.

vi) We reject the null if p-value <= 0.01.

vii) Here, we see the p-value>0.01 so we accept the null and conclude that the proportion of scrap is not more than 0.171255.

b) The critical value of the test obtained from the normal table is 2.326.

We reject the null if T>critical value. Here. T< 2.326 so we accept the null and conclude that the proportion of scrap is not more than 0.171255.

Thus we see for both critical value approach and p-value approach used in part a results in same answer.

c) The confidence bound is given by-

=

= (0.1225,0.3974)

We know that if the confidence interval doesnot contain the null hypothesis value the test is significant.

But here, the null hypothesis value lies between the confidence interval so we say that the test is not significant.