Answer:
a. Mean assembly time is estimated to fall between 15.6mins and 16.8mins with a 92% CI.
b.636 workers
Step-by-step explanation:
Let [tex]\mu[/tex] denote the mean assembly time in minutes.
Standard deviation,[tex]\sigma=3.6min[/tex]
To obtain a 92%CI for [tex]\mu[/tex] based on given parameters:[tex]n=120,[/tex] [tex]\bar X=16.2min[/tex],[tex]\alpha =1-0.92=0.08,\sigma=3.6[/tex]. [tex]\sigma[/tex] is known, therefore we use the [tex]z-[/tex]interval:
[tex]\bar X\pm z_0_._5_n\frac{\sigma}{\sqrt n}=16.2\pm 1.75\frac{3.6}{\sqrt 120}=16.2\pm0.575\\=[15.6,16.8][/tex]
Where[tex]z_0_._0_4=1.75[/tex], and is obtained from the z-interval tables.
b. A sample of size n needs to be completed based on the info:[tex]\alpha =0.08, \sigma=3.6, E=15sec=0.25min[/tex], E denotes the margin of error.
Therefore the sample size is calculated as:-
[tex]n=(\frac{z_\frac{\alpha}{z^\sigma}}{E})^2=(\frac{1.75\times3.6}{0.25})^2\\=636[/tex]
Hence, 636 workers should be involved in this study in order to have the mean assembly time required.
