Question

Question 2
A study of a large production facility was undertaken to determine the average time required to assemble a microchip. A random sample of 15 assemblies produced a sample mean of 12.2 minutes with a standard deviation of 2.4 minutes. Assuming that the assembly times are normally distributed, construct a 95% confidence interval to estimate the population mean assembly time. Use the following template to answer the question.
I. Parameter of interest: …………………………………….
II. Point estimator: ………………..
III. Sampling distribution of the point estimator: ………………..
IV. Specify the formula for the 95% confidence interval estimator for the parameter: …………………………………………………………….
V. Perform the necessary calculations and write down the lower and upper limits of the confidence interval:
VI. Interpret the calculated confidence interval.

Answer 1

1) population mean assembly time

2) Point estimator: sample mean of 12.2 minutes

III.Standard Error , SE = s/√n =   2.4000   / √    15   =   0.6197

Sampling distribution of the point estimator:  normally distributed , N ~(12.2 , 0.6197)

IV. confidence interval is x̅ ± critical value*s/√n

Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   14          
't value='   tα/2=   2.145   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   2.4000   / √   15   =   0.6197
margin of error , E=t*SE =   2.1448   *   0.6197   =   1.3291
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    12.20   -   1.329076   =   10.8709
Interval Upper Limit = x̅ + E =    12.20   -   1.329076   =   13.5291
95%   confidence interval is (   10.87   < µ <   13.53   )

VI) there is 95% confidence that population mean assembly time lies within confidence interval