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victoriadorvilu

21.03.2023

Arts

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 $\mu$ denote the mean assembly time in minutes.

Standard deviation, $\sigma=3.6min$

To obtain a 92%CI for $\mu$ based on given parameters: n=120, $\bar X=16.2min$ , $\alpha =1-0.92=0.08,\sigma=3.6$ . $\sigma$ is known, therefore we use the interval:

$\bar X\pm z_0_._5_n\frac{\sigma}{\sqrt n}=16.2\pm 1.75\frac{3.6}{\sqrt 120}=16.2\pm0.575\\=[15.6,16.8]$

Where z_0_._0_4=1.75 , and is obtained from the z-interval tables.

b. A sample of size n needs to be completed based on the info: $\alpha =0.08, \sigma=3.6, E=15sec=0.25min$ , E denotes the margin of error.

Therefore the sample size is calculated as:-

$n=(\frac{z_\frac{\alpha}{z^\sigma}}{E})^2=(\frac{1.75\times3.6}{0.25})^2\\=636$

Hence, 636 workers should be involved in this study in order to have the mean assembly time required.

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