A manufacturer wants to assess whether or not rods are being constructed appropr

Question

A manufacturer wants to assess whether or not rods are being constructed appropriately, where the diameter of the rods is supposed to be 1.0 cm, and the variation in the diameters is known to be distributed N(mu, 0.1). The manufacturer is willing to tolerate a deviation of the population mean from this value of no more than 0.1 cm, i.e., if the population mean is within the interval 1.0plusminus0.1 cm, then the manufacturing process is performing correctly. A sample n = 500 rods is taken, and the average diameter of these rods is found to be x = 1.05 cm, with s^2 = 0.083 cm^2. Are these results statistically significant? Are the results practically significant? Justify your answers.

Explanation / Answer

In order to see whether the manufacturing process for the given sample performs perfectly, let us first calculate the range for the sample mean.

The expected range of diameter of the rods is 1.00 =-0.1cm or in other words [0.90,1.10]

This implies we need to calculate the confidence interval of sample mean

give:

Sample size = 500 rods

xbar = 1.05cm

s^2 = 0.083

To find confidence interval we need critical value of Z for a given value of significance.

Since nothing is mentioned about the significance we will consider the default value or the most commonly used value of significance ie. 95%.

The Z critical value for significance level 95% is 1.96

The formula to be used to calculate the confidence interval of sample mean is as follows:

[xbar - Zcritical * s/root n, xbar + Zcritical * s/root n]

on plugging in all the values in the formula we get

[1.05 - 1.96* (0.083/500), 1.05 + 1.96 * (0.083/500)]

[1.05 - 1.96* 0.01288, 1.05 + 1.96 * 0.01288]

[1.05 - 0.02524, 1.05 + 0.02524]

[1.02476, 1.07524]

The sample range falls within the expected range hence indicating the correct performance.

Statistical significance of results: If such a sample is selected a number of times, then 95% of such samples will have the true population mean in the calculated confidence itnterval.

To check the practical significance:

Let us first calculate the test statistic Z for th given sample:

Z = (xbar - mean)/(s/n)

mean of population is 1.00

xbar = 1.05

s = root of 0.083

n = 500

Hence Z = (1.05-1.00)/(0.083/500)

Z = 0.05/0.01288

Z = 3.882

The P value of 3.882 an be found using the Z table, or online Pvalue calculator for Z, or making use of excel formula "normsdist(3.882) , we get the P value as 0.9999, which implies the probability of occurance of the such samples that show perfect performance.

Hence the results are practically significant.

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