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?

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".

[tex]\bar X=1.3[/tex] represent the sample mean

[tex]\mu[/tex] 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:

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)

In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:

[tex]df=n-1=9-1=8[/tex]

Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], 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 [tex]t_{\alpha/2}=2.306[/tex]

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

[tex]1.30-2.306\frac{0.2}{\sqrt{9}}=1.146[/tex]

[tex]1.30+2.306\frac{0.2}{\sqrt{9}}=1.454[/tex]

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:

[tex] ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (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:

[tex]n=(\frac{z_{\alpha/2} \sigma}{ME})^2[/tex] (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 [tex]z_{\alpha/2}=2.326[/tex], replacing into formula (b) we got:

[tex]n=(\frac{2.326(0.2)}{0.065})^2 =51.22 \approx 52[/tex]

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