Question

We run a terminating simulation 20 replications to estimate the average throughput of a manufacturing system. Suppose that the observed throughputs from these 20 replications are 4, 5, 7, 5, 8, 7, 5, 9, 10, 5, 6, 7, 2, 11, 6, 3, 2, 11, 10, 7. a. (0.1 point) What is the 99% t-confidence interval for the mean throughput

Answer 1

`CI =6.5 \± 1.09`

Explanation:

Given

`Data: 4, 5, 7, 5, 8, 7, 5, 9, 10, 5, 6, 7, 2, 11, 6, 3, 2, 11, 10, 7`

Required

Determine the 99% t-confidence interval

First, calculate the mean

`\bar x =(\sum x)/(n)`

`\bar x =(4+ 5+ 7+ 5+ 8+ 7+ 5+ 9+10+ 5+ 6+ 7+ 2+ 11+ 6+ 3+ 2+ 11+ 10+ 7)/(20)`

`\bar x =(130)/(20)`

`\bar x =6.5`

Calculate the standard deviation

`\sigma = \sqrt{(\sum(x - \bar x))/(n)`

`\sigma =\sqrt{((4-6.5)^2+ (5-6.5)^2+ ........+ (2-6.5)^2+ (11-6.5)^2+ (10-6.5)^2+ (7-6.5)^2)/(20)}`

`\sigma =\sqrt{(143)/(20)}`

`\sigma =√(7.15)`

`\sigma =2.67`

The z score for 99% confidence interval is: 2.576

The confidence interval is calculated as:

`CI =\bar x \± z * (\sigma)/(\sqrt n)`

`CI =6.5 \± 2.576 * \frac{2.67}{\sqrt {40}}`

`CI =6.5 \± 2.576 * (2.67)/(6.32)}`

`CI =6.5 \± 1.09`