Question

For 113 consecutive days, a process engineer has measured the temperature of champagne bottles as they are made ready for serving. Each day, she took a sample of 9 bottles. The average across all 1017 bottles (113 days, 9 bottles per day) was 60 degrees Fahrenheit. The standard deviation across all bottles was 0.5 degrees. (Round your intermediate calculations and final answer to 4 decimal places.) When constructing an X-bar chart, what would be the upper control limit

Answer 1

upper control limit
`UCL=7.1632`

Concept :

`UCL=\bar{\bar{x}}+A_2\bar{R}` where
`A_2=0.34`

`\bar{\bar{x}}=\frac{\bar{x}}{n}`

`S=\frac{\bar{R}}{d}\Rightarrow \bar{R}=S* d` where
`S=0.5, d=2.97`

Given :

She took a sample of per day
`(n)` is
`9` bottles.

The standard deviation
`S`of all bottles is
`0.5` degree.

Average across all
`1017` bottles
`\bar{x}`are
`60` degree Fahrenheit.

To find :

The value of the upper control limit (UCL)

Explanation :

Here, firstly we find the value of
`\bar{\bar{x}}` .

`\because\bar{\bar{x}}=\frac{\bar{x}}{n}`

`\therefore \bar{\bar{x}}=(60)/(9)`

`\Rightarrow \bar{\bar{x}}=6.66` and
`\bar{R}=S* d`

`\Rightarrow \bar{R}=0.5*2.97=1.48`

`\Rightarrow \bar{R}=1.48`

Now,
`UCL=\bar{\bar{x}}+A_2\bar{R}`

`\Rightarrow UCL=6.66+0.34*1.48`

`\Rightarrow UCL=6.66+0.5032`

`\Rightarrow UCL=7.1632` degrees.