Question

7. Overhead Door (OD) Corporation's founder, C. G. Johnson, invented the upward-lifting garage door in 1921 and the electric garage door opener in 1926. Since then OD has been a leading supplier of commercial, industrial, and residential garage doors sold through a nationwide network of more than 450 authorized distributors. They have built a solid reputation as a premier door supplier, commanding 15 % share of the market. Suppose that customers assess door quality first in terms of the ease of operation, followed by its durability. The quality improvement team (QIT) might then assign an engineering team to determine the factors that contribute to these two main problems. Smooth operation of a garage door is a critical quality characteristic that affects both problems: If a door is too heavy, it's difficult and unsafe to balance and operate; if it's too light, it tends to buckle and break down frequently or may not close properly. Suppose the design engineers determine that a standard garage door should weigh a minimum of 75.5 kg. and a maximum of 85.5 kg., which thus specifies its design quality specification. QIT is inspecting if there is evidence that the percentage of defective doors does not exceed 10%. Suppose the QIT decides to collect data on the actual weights of 60 standard garage doors sampled randomly from their monthly production of almost 2,000 doors. See the tables on the next page. 7.1 (2 point) What is the sample defective rate ? 7.2 (2 points) Formulate and test an appropriate set of hypotheses to determine if the machine can be qualified. Use a = 0.05. Find the P-value. 7.3 (2 points) Is there evidence to support the claim that a standard garage door weigh 80 kg.? What is the P-value? Use a = 0.01. 7.4 (2 points) What is the 99% confidence interval?

Answer 1

(7.1)$\bar{p}$=x/n=10/60=0.1667

There are 7 observations below the 75.5 and 3 observations above 85.5, so total x=10 observations are defective out of n=60

(7.2)here we want to test the null hypothesis H0:$\bar{P}$=0.10 and alternate hypothesis Ha:$\bar{P}$>0.10

here we use z-test and statistic z=($\bar{p}$-$\bar{P}$)/SE($\bar{p}$)=(0.1667-0.10)/0.0387=1.724

SE($\bar{p}$)=sqrt($\bar{P}$*(1-$\bar{P}$)/n)=sqrt(0.10*(1-0.10)/60)=0.0387

this is one tailed test and one-tailed critical z(0.05)=1.645 is less than calculated z=1.724, so reject the null hypothesis and conclude that machine couldn't qualify.

one tailed p-value=0.0424 is less than alpha=0.05, so we reject the null hypothesis and conclude that machine couldn't qualify.

(7.3) here we want to test the null hypothesis H0:$\mu$=80 and alternate hypothesis Ha:
_{u #80}

test statistic t=($\bar{x}$-$\mu$)/(s/sqrt(n))=(80.2-80)/(3.7678/sqrtt(60))=0.4112with n-1=60-1=59 df

the two tailed critical t=2.662 is more than calculated t=0.411, so we fail to reject H0 and conclude that mean weight is 80

(7.4)

n=	60
sample mean=	80.2
s=	3.7678

(1- α)*100% confidence interval for population mean=sample mean±t(α/2,n-1)*s/sqrt(n)

99% confidence interval for population mean=mean±margin of error=mean±t(0.01/2, n-1)*s/sqrt(n)=80.2±1.295=(78.905,81.495)

	t-value	margin of error	lower limit	upper limit
99% confidence interval	2.662	1.295	78.905	81.495

observations arranged in ascending order

72	78	82
73	79	83
74	79	83
74	79	83
75	79	83
75	80	83
75	80	84
76	80	84
76	80	84
76	81	84
76	81	84
76	81	84
76	81	84
76	81	85
77	82	85
77	82	85
78	82	85
78	82	86
78	82	86
78	82	88