Question

n a packing plant, one of the machines packs jars into a box. A sales rep for a packing machine manufacturer comes into the plant saying that a new machine he is selling will pack the jars faster than the old machine. To test this claim, each machine is timed for how long it takes to pack 10 cartons of jars at randomly chosen times. Given a 99% confidence interval of (-8.6, -3.08) for the true difference in average times to pack the jars (old machine - new machine), what can you conclude from this interval?

Answer 1

`(\bar X_(old) -\bar X_(new)) + t_(\alpha/2) \sqrt{(s^2_(old))/(n_(old)) +(s^2_(new))/(n_(new))}`

After replace we got for the interval:

`(-8.6,-3.08)`

For this case since the interval have just negative values then we can conclude that at 1% of significance the mean for the old case is lower than the mean for the new machine and for this case we can conclude that the new machine reduce the times to pack the jars.

Explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

For this case we assume the following:

`\bar X_(old)` represent the man for old case

`\bar X_(new)` represent the man for new case

`s_(old)` deviation for the old case

`s_(new)` deviation for the new case

For this case we want to find a confidence interval for the difference of means and they can use the following formula:

`(\bar X_(old) -\bar X_(new)) + t_(\alpha/2) \sqrt{(s^2_(old))/(n_(old)) +(s^2_(new))/(n_(new))}`

After replace we got for the interval:

`(-8.6,-3.08)`

For this case since the interval have just negative values then we can conclude that at 1% of significance the mean for the old case is lower than the mean for the new machine and for this case we can conclude that the new machine reduce the times to pack the jars.