Diamond size is measured in carats. For ladies diamond rings, it is known the me

Question

Diamond size is measured in carats. For ladies diamond rings, it is known the mean diamond size is 0.2 carats with a standard deviation of 0.057 carats. A company collected a random sample of 36 ladies diamond rings from all diamond companies and found these diamonds had a mean size of 0.203 carats. What is the population? What is the population parameter that we are interested in for this problem? What is the sample? What is the sample statistic? Check the conditions needed for having a Normal sampling distribution. Based on Question e), what is the sampling distribution of the sample statistic? If we repeated the sampling process with the same sample size, what proportion of the samples will have the mean diamond size between 0.202 and 0.204 carats?

Explanation / Answer

a)

All ladies' diamond rings

b)

Mean size of all ladies' diamond rings

c)

a random sample of 36 ladies diamond rings

d)

Xbar = 0.203 carats

e)

As n > 30, and the sample is a simple random sample, then we have a normal sampling distirbution.

f)

It has a mean of 0.2 carats, and standard deviaiton of

s(Xbar) = s/sqrt(n) = 0.057/sqrt(36) = 0.0095 [ANSWER]

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g)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    0.202      
x2 = upper bound =    0.204      
u = mean =    0.2      
n = sample size =    36      
s = standard deviation =    0.057      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    0.210526316      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    0.421052632      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.583371543      
P(z < z2) =    0.663141675      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.079770133   [ANSWER]  

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