To calculate the value of the test statistic, we need to use the formula for the z-test statistic, which is given as:
\[ z = \frac{\bar{X} - \mu}{\sigma / \sqrt{n}} \]
Where:
- \(\bar{X}\) is the sample mean (which in this case is 55).
- \(\mu\) is the population mean (which in this case is 60).
- \(\sigma\) is the sample standard deviation (which is given as 20).
- \(n\) is the sample size (which is 100).
Now, substituting the values into the formula:
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The difference between the sample mean and the population mean is: \[ \bar{X} - \mu = 55 - 60 = -5 \]
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The standard error of the mean is calculated as: \[ \sigma / \sqrt{n} = 20 / \sqrt{100} = 20 / 10 = 2 \]
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Now, we can calculate the z-value: \[ z = \frac{-5}{2} = -2.5 \]
Now let's evaluate the options provided:
A) \(z = \frac{(55 - 60)}{20}\)
B) \(z = \frac{(55 - 60)}{(20/100)}\)
C) \(z = \frac{(55 - 60)}{(20/100^2)}\)
D) \(z = \frac{(55 - 60)}{(20/10)}\)
- Option A: Incorrect, since it's not dividing by the standard error.
- Option B: Incorrect, it should be the standard error \(20/\sqrt{100}\), not \(20/100\).
- Option C: Incorrect, the denominator should not have \(100^2\).
- Option D: Correct, as it uses the right standard error \((20/10)\).
Based on the analysis and calculations, the correct expression used for the z-value is represented in Option D, \(z = \frac{(55 - 60)}{(20/10)}\), even though the actual value of \(z\) after computation is not required. The method confirms your approach is valid.
1 answer