A certain company has a process producing measurements that are i.i.d. and norma

Question

A certain company has a process producing measurements that are i.i.d. and normal. Suppose that the process is considered to be in control as long as the population standard deviation of the measurements is no larger than 0.015. For the questions below, suppose that the process is just barely under control, so sigma = 0.015. (You will probably need to find a chi^2 distribution calculator!) (a) Determine the probability that the sample standard deviation, from a random sample of 10 measurements (and based on a divisor of n - 1), will exceed 0.02. (b) Repeat part (a) with a sample size of 50 measurements. (c) Comment on how this probability has changed, and explain why this makes sense. (d) With a sample size of 10, determine how large would the sample standard deviation have to be in order for the probability to be only 0.01 of obtaining such a large sample standard deviation if the process were (just barely) under control. (e) Repeat part (d) for sample sizes of 50. (f) Comment on how these values (in (d) and (e)) compare, and explain why this makes sense.

Explanation / Answer

This example follow a ji square distribution:

Part a)

n= 10

P (standard deviation > 0.02)..... To calculate that's you need to calculate X2   distribution applying this formula:

X2 = (n-1)*S2 / (standard deviation)2   = (10 -1) * (0.015)2 / (0.02)2 = 5.06

The value of 5.06 is searched within the table of critics value of ji square distribution in row 9 (n-1) degrees of freedom and it is found that this value corresponds to an area to the right between 0.80 y 0.90.

So the P (standard deviation > 0.02) will be between 0.8 and 0.9, because we do not found a exactly value that corresponds to 5.06 in the tables of ji square distribution.

Part b)

n= 50

P (standard deviation > 0.02)..... To calculate that's you need to calculate X2   distribution applying this formula:

X2 = (n-1)*S2 / (standard deviation)2   = (50 -1) * (0.015)2 / (0.02)2 = 27.56

The value of 27.56 is searched within the table of critics value of ji square distribution in row 49 (n-1) degrees of freedom and it is found that this value corresponds to an area to the right between 0.975 y 0.995.

So the P (standard deviation > 0.02) with n = 50 will be between 0.975 and 0.995, because we do not found a exactly value that corresponds to 27.65 in the tables of ji square distribution.

Part c)

In this case the probability increased.

Part d)

In this case contrary to the first two cases the problem ask the standard deviation and give us as information the the probability. Using the tables of ji square distribution we can obtain X2 .

9 grades of freedon and 0.01 of probability X2 = 21.66

with this value we can calculate the standard deviation solving this equation X2 = (n-1)*S2 / (standard deviation)2 .

To solve this equation the standard deviation for P = 0.01 and n = 10 is equal to 0.0096

Part e)

This is the same case of part d. but n = 50

49 grades of freedon and 0.01 of probability X2 = 74.919

with this value we can calculate the standard deviation solving this equation X2 = (n-1)*S2 / (standard deviation)2 .

To solve this equation the standard deviation for P = 0.01 and n = 50 is equal to 0.0121

Part F)

The standar deviation increase when the sample size increase of 50.

as we increase the deviation increased to approach thevalue of the reference deviation for this example

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