Question

1. a sample of ( n = 64 ) scores has a mean of ( m = 68 ). assuming that the population mean is ( mu = 60 ), find the ( z )-score for this sample:

a. if it was obtained from a population with ( sigma = 16 )
b. if it was obtained from a population with ( sigma = 32 )
c. if it was obtained from a population with ( sigma = 48 )

Explanation:

Step1: Recall the formula for the z - score of a sample mean

The formula for the z - score of a sample mean is \(z=\frac{M-\mu}{\frac{\sigma}{\sqrt{n}}}\), where \(M\) is the sample mean, \(\mu\) is the population mean, \(\sigma\) is the population standard deviation, and \(n\) is the sample size. Given \(n = 64\), \(M=68\), \(\mu = 60\).

Step2: Calculate for part (a)

When \(\sigma = 16\)
First, calculate \(\frac{\sigma}{\sqrt{n}}=\frac{16}{\sqrt{64}}=\frac{16}{8}=2\)
Then, \(z=\frac{M - \mu}{\frac{\sigma}{\sqrt{n}}}=\frac{68 - 60}{2}=\frac{8}{2}=4\)

Step3: Calculate for part (b)

When \(\sigma = 32\)
First, calculate \(\frac{\sigma}{\sqrt{n}}=\frac{32}{\sqrt{64}}=\frac{32}{8}=4\)
Then, \(z=\frac{M-\mu}{\frac{\sigma}{\sqrt{n}}}=\frac{68 - 60}{4}=\frac{8}{4}=2\)

Step4: Calculate for part (c)

When \(\sigma = 48\)
First, calculate \(\frac{\sigma}{\sqrt{n}}=\frac{48}{\sqrt{64}}=\frac{48}{8}=6\)
Then, \(z=\frac{M-\mu}{\frac{\sigma}{\sqrt{n}}}=\frac{68 - 60}{6}=\frac{8}{6}=\frac{4}{3}\approx1.33\)

Answer:

a. \(z = 4\)
b. \(z = 2\)
c. \(z\approx1.33\)