Question

A cola-dispensing machine is set to dispense 11 ounces of cola per cup, with a standard deviation of 1.0 ounce. The manufacturer of the machine would like to set the control limit in such a way that, for samples of 35, 5% of the sample means will be greater than the upper control limit, and 5% of the sample means will be less than the lower control limit.Required:a. At what value should the control limit be set?b. If the population mean shifts to 10.7, what is the probability that the change will be detected?c. If the population mean shifts to 11.7, what is the probability that the change will be detected?

Answer 1

a. the control limits should be set at (10.72, 11.28)

b.
`\mathbf{P(10.72<x<11.28) = 0.4526}`

c.
`\mathbf{P(10.72<x<11.28) = 0.0065}`

Explanation:

Given that:

population mean μ = 11

standard deviation
`\sigma` = 1.0

sample size n = 35

5% of the sample means will be greater than the upper control limit, and 5% of the sample means will be less than the lower control limit.

Therefore, level of significance ∝ = 0.05+0.05 = 0.10

Critical value for
`z_(1-\alpha/2) =z_(1-0.10 /2)`

`\implies z_(1-0.05) = z_(0.95)`

Using the EXCEL FORMULA: = NORMSINV (0.95)

z = 1.64

The lower control limit and the upper control limit can be determined by using the respective formulas:

Lower control limit =
`\mathtt{\mu - z_(1-\alpha/2) * (\sigma)/(√(n))}`

Upper control limit =
`\mathtt{\mu + z_(1-\alpha/2) * (\sigma)/(√(n))}`

For the lower control limit =
`11-1.64 * (1.0)/(√(35))`

For the lower control limit =
`11-0.27721`

For the lower control limit = 10.72279

For the lower control limit
`\simeq` 10.72

For the upper control limit =
`11+1.64 * (1.0)/(√(35))`

For the upper control limit = 11 + 0.27721

For the upper control limit = 11.27721

For the upper control limit
`\simeq` 11.28

Therefore , the control limits should be set at (10.72, 11.28)

b. If the population mean shifts to 10.7, what is the probability that the change will be detected?

i.e

`\mathtt{P(10.72<x<11.28) = P( (X- \mu)/((\sigma)/(√(n)))<z < (X- \mu)/((\sigma)/(√(n)))})`

`\mathtt{P(10.72<x<11.28) = P( (10.72- 10.7)/((1.0)/(√(35)))<z < (11.28- 10.7)/((1.0)/(√(35)))})`

`\mathtt{P(10.72<x<11.28) = P( (0.02)/((1.0)/(5.916))<z < (0.58)/((1.0)/(5.916))})`

`\mathtt{P(10.72<x<11.28) = P(0.1183<z < 3.4313})`

`\mathtt{P(10.72<x<11.28) = P(z< 3.4313) - P(z< 0.1183) }`

Using the EXCEL FORMULA: = NORMSDIST (3.4313) - NORMSDIST (0.118 ); we have:

`\mathbf{P(10.72<x<11.28) = 0.4526}`

c If the population mean shifts to 11.7, what is the probability that the change will be detected?

i.e

`\mathtt{P(10.72<x<11.28) = P( (X- \mu)/((\sigma)/(√(n)))<z < (X- \mu)/((\sigma)/(√(n)))})`

`\mathtt{P(10.72<x<11.28) = P( (10.72- 11.7)/((1.0)/(√(35)))<z < (11.28- 11.7)/((1.0)/(√(35)))})`

`\mathtt{P(10.72<x<11.28) = P( (-0.98)/((1.0)/(5.916))<z < (-0.42)/((1.0)/(5.916))})`

`\mathtt{P(10.72<x<11.28) = P(-5.7978<z < -2.48472})`

`\mathtt{P(10.72<x<11.28) = P(z< -2.48472) - P(z< -5.7978) }`

Using the EXCEL FORMULA: = NORMSDIST (-2.48472) - NORMSDIST (-5.7978); we have:

`\mathbf{P(10.72<x<11.28) = 0.0065}`