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Answer:
Step-by-step 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".
[tex]\bar X[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean (variable of interest)
s represent the sample standard deviation
n represent the sample size
Solution to the problem
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 =0.1 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 95.5% of confidence interval now can be founded using the normal distribution. And in excel we can use this formula to find it:"=-NORM.INV(0.023;0;1)", and we got [tex]z_{\alpha/2}=2.004[/tex], replacing into formula (b) we got:
[tex]n=(\frac{2.004(0.9)}{0.1})^2 =3.25 \approx 4[/tex]
So the answer for this case would be n=4 rounded up to the nearest integer
By using formula of margin error we got that total 4 observations should a time study analyst plan for in an operation that has a standard deviation of .9 minutes per piece if the goal is to estimate the mean time per piece to within .1 minute with a confidence of 95.5 percent
What is standard deviation?
Standard deviation is a number used to tell how measurements for a group are spread out from the average (mean), or expected value.
Here given that
standard deviation = 0.9
confidence level= 95.5%
Margin error= 0.1
We know that margin error can be written as:
[tex]$M E=z_{\alpha / 2} \frac{\sigma}{\sqrt{n}} \quad$[/tex]
Here
[tex]M E=\text{margin error}\\\\\sigma= \text{standard deviation}\\\\n=\text{Number of observation}[/tex]
Now we can calculate number of observations by using z-score table as
[tex]$n=\left(\frac{2.004(0.9)}{0.1}\right)^{2}=3.25 \approx 4$[/tex]
By using formula of margin error we got that total 4 observations should a time study analyst plan for in an operation that has a standard deviation of .9 minutes per piece if the goal is to estimate the mean time per piece to within .1 minute with a confidence of 95.5 percent
To learn more about standard deviation visit :https://brainly.com/question/12402189
