Question

A total of 9 cycles have been observed during a direct time study. The mean for the largest element time = 1.30 min, and the corresponding sample standard deviation s = 0.20 min. (a) Based on these data, what is the 95% confidence interval on the 1.30 min element time? (b) If the analyst wants to be 98% confident that the mean of the sample was within 5% of the true mean, how many more observations should be taken?

Answer 1

a)
`1.30-2.306(0.2)/(√(9))=1.146`

`1.30+2.306(0.2)/(√(9))=1.454`

So on this case the 95% confidence interval would be given by (1.146;1.454)

b)
`n=((2.326(0.2))/(0.065))^2 =51.22 \approx 52`

So the answer for this case would be n=52 rounded up to the nearest integer

Explanation:

Previous concepts

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".

`\bar X=1.3` represent the sample mean

`\mu` population mean (variable of interest)

s=0.2 represent the sample standard deviation

n=9 represent the sample size

Part a

The confidence interval for the mean is given by the following formula:

`\bar X \pm t_(\alpha/2)(s)/(√(n))` (1)

In order to calculate the critical value
`t_(\alpha/2)` we need to find first the degrees of freedom, given by:

`df=n-1=9-1=8`

Since the Confidence is 0.95 or 95%, the value of
`\alpha=0.05` and
`\alpha/2 =0.025`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,8)".And we see that
`t_(\alpha/2)=2.306`

Now we have everything in order to replace into formula (1):

`1.30-2.306(0.2)/(√(9))=1.146`

`1.30+2.306(0.2)/(√(9))=1.454`

So on this case the 95% confidence interval would be given by (1.146;1.454) Part b

The margin of error is given by this formula:

`ME=z_(\alpha/2)(\sigma)/(√(n))` (a)

And on this case we have that ME =0.05*1.3= 0.065 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

`n=((z_(\alpha/2) \sigma)/(ME))^2` (b)

The critical value for 98% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.01;0;1)", and we got
`z_(\alpha/2)=2.326`, replacing into formula (b) we got:

`n=((2.326(0.2))/(0.065))^2 =51.22 \approx 52`

So the answer for this case would be n=52 rounded up to the nearest integer